I just finished taking my first (and maybe only) functional analysis class spring quarter! and with that, the 2024-25 season of core classes comes to an end. I am thankfully now freed of any more core class obligations, which is a huge huge relief.

Jumping straight into a functional analysis course having not taken the preceding measure class here at UW was a little scary, especially my knowledge of measure theory was nearly two and a half years old. I also had felt that in general, I hadn’t really absorbed the graduate real analysis class I took in undergrad (that I did relatively poorly in) and had subsequently mostly purged it from my brain. Suffice it to say I was fairly on edge about not knowing certain measure theory facts that I felt I should have learned. I should really have given myself a bit more grace, especially since the texts that the courses used were different (Folland vs. Stein-Shakarchi) and had covered slightly different things, to no fault of my own.

Luckily, my knowledge gaps weren’t too much of a problem, but a couple things I missed were various theorems about decomposing measures, something that I just never saw (in the case of Hahn or Jordan decomposition) or just didn’t learn (in the case of Radon-Nikodym and the corresponding decomposition). I had always meant at some point to learn what some of this decomposition business was, and at a certain point this quarter I was forced to in order to do a problem on a problem set. I’ve since decided to take the crash coursing of measure theory stuff that I had to do this quarter and turn it into a bit of a post-quarter learning experience here, so here goes…

(essentially most of what follows is exposition from Folland’s Real Analysis (2nd ed.), Chapter 3, sections 1-2.)

hold on, do YOU remember measure theory?

if you don’t remember what a measure is, you might want to unroll this

For a measurable space \(X\), the eponymous measure \(\mu\) is not one that is defined on the entirety of the power set \(\mc P(X)\), but only on a \(\sigma\)-algebra \(\mc M\) over \(X\), i.e. a subset of the power set closed under complementation and countable union/intersection. I refer to this collection \(\mc M\) as the collection of measurable sets, since it forms the domain of \(\mu : \mc M \to [0, \infty]\) (and can take on the infinite value). A measure satisfies the following three conditions:

  • For any \(E \in \mc M\), \(\mu(E) \geq 0\) (true by definition, but worth reiterating for effect).
  • \[\mu(\nullset) = 0\]
  • Countable additivity: For any mutually disjoint collection of subsets \(\set{E_i}_{i=1}^\infty \subseteq \mc M\), \(\mu(\bigcup_{i=1}^\infty E_i) = \sum_{i=1}^\infty \mu(E_i)\).

Measures are our abstractions of “area” or “volume” functions – they assign a notion of area/volume to subsets of a space, but there are issues with being able to do this to all subsets of a space if we have a sufficiently large (uncountable) space from axiom of choice nonsense.

We can extend the notion of a measure to a signed measure \(\nu : \mc M \to [-\infty, \infty]\) if we drop the condition that \(\nu\) need be non-negative and instead assert that \(\nu\) assumes at most one of the values \(\pm \infty\). Signed measures behave essentially the same as normal measures, but with signs they form a real vector space under pointwise addition. For a signed measure \(\nu\), a measurable set \(E\) is said to be positive if for every \(F \subseteq E\) measurable we have \(\nu(F) \geq 0\) (and similarly negative if \(\nu(F) \leq 0\)). We assume all measures are positive unless otherwise specified.

We can write down signed measures with extended-valued measurable functions \(f : X \to [-\infty, \infty]\) (with respect to some measure \(\mu\)). Here, we require that for the positive and negative parts of \(f\), \(f^+\) and \(f^-\), respectively, at most one of \(\int f^+ \, d\mu\) and \(\int f^- \, d\mu\) is infinite. Such an \(f\) defines a measure from another one \(\mu\) by \(\nu(E) = \int_E f \, d\mu\). Notationally, we can express this by saying \(d\nu = f \, d\mu\).

Finally, a few notions about how measures can relate to each other:

  • Two signed measures \(\mu, \nu\) are called mutually singular (or we say \(\nu\) is singular with respect to \(\mu\), or vice versa) if there is a decomposition \(X = E \sqcup F\) with \(\mu(E) = \nu(F) = 0\). This is denoted \(\mu \perp \nu\), and one can think of this as saying “\(\mu\) and \(\nu\) live on disjoint sets.”
  • For a (signed) measure \(\nu\) and a measure \(\mu\), \(\nu\) is absolutely continuous with respect to \(\mu\) if for every measurable \(E\) with \(\mu(E) = 0\) we have \(\nu(E) = 0\). This is related to the notion of absolute continuity for real-valued functions1.

breaking stuff

Before we get to some of the interesting ways we can break up a measure, we need to get some obvious-looking stuff out of the way. Before this quarter, I never learned anything about how to decompose a signed measure, so I figure it’s worth talking a bit about the basics before talking about more high-power stuff.

First, from the “one-sided” finitenses of a signed mesaure, I’d like for a signed measure to break up nicely into a positive and a negative part, so we can view them as a difference of two positive measures. This is essentially what Hahn decomposition and Jordan decomposition say.

First, Hahn decomposition basically says that signed measures are positive on one part of the space and negative on another:

Theorem. (Hahn Decomposition) If \(\nu\) is a signed measure, then there is a positive set \(P\) and a negative set \(N\) such that \(X = P \sqcup N\). This is unique up to null sets – if \(X = P' \sqcup N'\) for \(P'\) and \(N'\) positive and negative, respectively, then \(P \triangle P'\) and \(N \triangle N'\) are null sets (for \(\nu\)).

Proof. Assuming without loss of generality that \(\nu\) does not achieve \(+\infty\), we first construct \(P\) and \(N\). Let \(M\) be \(\sup_{E \in \mc M \text{ positive}} \nu(E)\), and pick a sequence of positive sets \(\set{P_i}_{i=1}^\infty\) such that \(\nu(P_i) \to M\). Then we can let \(P = \bigcup_{i=1}^\infty P_i\) be our desired positive set, where by the continuity of measure we have that \(\nu(P) = \nu(\bigcup_{i=1}^\infty P_i) = \lim_{i \to \infty} \nu(P_i) = M\). We claim \(P\) and its complement \(N = X - P\) give our desired pair of sets.

We need to check now that \(N\) is negative. This is maybe a little more delicate than it might seem. We can first check that \(N\) contains no non-null positive sets – otherwise, if \(A \subseteq N\) is positive, then \(\nu(A \cup P) = \nu(A) + \nu(P) \leq \nu(P)\), and since \(\nu(A) \geq 0\), we must have that \(\nu(A) = 0\). However, there could still be some \(A \subseteq N\) with \(\nu(A) > 0\), which we have to be careful about. However, having such an \(A\) not be positive allows us to construct a descending filter of sets of increasing measure. Note that there must be some subset \(B \subseteq A\) with \(\nu(B) < 0\)$ if \(A\) is not positive, but then \(\nu(A - B) + \nu(B) = \nu(A)\) gives that \(\nu(A - B) = \nu(A) - \nu(B) > \nu(A)\).

Iterating this construction assuming \(N\) is not negative, we construct a sequence of subsets \(\set{A_i}_{i=1}^\infty\) and integers \(\set{n_i}_{i=1}^\infty\). First, pick \(n_1\) to be the smallest integer for which there exists a set \(A_1 \subseteq N\) with \(\nu(A_1) > \frac 1{n_1}\), and inductively pick \(n_i\) to be the smallest integer for which there exists a set \(A_i \subseteq A_{i-1}\) with \(\nu(A_i) > \nu(A_{i-1}) + \frac 1{n_i}\) by the above. Now take \(A = \bigcap_{i=1}^\infty A_i\), and we show that such a set is problematic. Since \(\infty > \nu(A) = \lim_{i \to \infty} \nu(A_i) > \sum_{i=1}^\infty \frac 1{n_i}\). This infinite sum converges, so in particular we must have \(\lim_{i\to \infty} n_i = \infty\). But then using the above observation again, we can find some \(B \subseteq A\) with \(\nu(B) > \nu(A) + \frac 1n\). But then since \(\set{n_i}_{i=1}^\infty\) shoots off to infinity, \(n < n_I\) for some \(I\) sufficiently large and \(B \subseteq A_{I-1}\) for such an \(I\), which means that we should have picked \(B\) to take out of \(A\) at some point in our construction. This is a contradiction, so \(N\) can’t be not negative, and therefore it must be negative.

A remark here: This feels like one of those really weird analysis arguments where you make an assumption, do a bunch of stuff, and then throw your hands up and go “whoops, that was bad! don’t do that!” Unfortunately this doesn’t really tell me why the opposite assumption must be true (law of excluded middle and all that) but I do think this argument is getting a lot of juice of the maximality of \(P\).

We have to do uniqueness now, but this is easy – if we also have \(X = P' \sqcup N'\), then \(P - P' \subseteq P\) and \(P - P' \subseteq N'\), so \(P - P'\) is positive and negative and hence is null. The same holds for \(P' - P\), and now we’re done.

From this, Jordan decomposition allows you to manifestly write down a signed measure as a difference of two positive ones:

Theorem. (Jordan Decomposition) If \(\nu\) is a signed measure, there are unique positive measures \(\nu^+\) and \(\nu^-\) with \(\nu = \nu^+ - \nu^-\) and \(\nu^+ \perp \nu^-\).

Note that we actually get uniqueness on the nose here, instead of uniqueness up to a set of measure zero. We want to leverage Hahn decomposition to do this.

Proof. Let \(X = P \sqcup N\) be a Hahn decomposition for \(\nu\), and we claim \(\nu^+(E) = \nu(P \cap E)\) and \(\nu^-(E) = - \nu(E \cap N)\) works. Clearly \(\nu^+(E) + \nu^-(E) = \nu(E)\), and \(\nu^+ \perp \nu^-\) with \(P\) and \(N\) as our partition of \(X\). The uniqueness is more interesting – suppose we have \(\nu = \mu^+ - \mu^-\) with \(\mu^+ \perp \mu^-\). We leverage the uniqueness we have with the Hahn decomposition here. Suppose \(\mu^+ \perp \mu^-\) with \(E \sqcup F = X\), \(E \cap F = \nullset\), \(\mu^+(F) = \mu^-(E) = 0\). We claim that \(E \sqcup F\) is another Hahn decomposition for \(X\) (which we can see more or less by construction). By the uniqueness, then, we have that \(\nu(P \triangle E) = 0\). This shows us that \(\mu^+ = \nu^+\), since for any measurable set \(A \in \mc M\), we have that \(\mu^+(A) = \mu^+(A \cap E) + \mu^+(A - E)\), and since \(A - E \subseteq F\) with \(\mu^+\) a positive measure, \(\mu^+(A - E) = 0\). Since \(\mu^-(A \cap E) = 0\), we see that this is also equal to \(\mu^+(A \cap E) - \mu^-(A \cap E) = \nu(A \cap E)\). Now, using the fact that \(\nu(P \triangle E) = 0\), \(\nu(A \cap E) = \nu(A \cap P) = \nu^+(A)\), as desired. This shows \(\nu^+\) is unique, and similarly so is \(\nu^-\).

If \(\nu\) is a signed measure, we then get the positive and negative variations \(\nu^+\) and \(\nu^-\) out of the Jordan decomposition. As an aside, we can now define the total variation of a signed measure \(\nu = \nu^+ - \nu^-\) to be \(\abs \nu = \nu^+ + \nu^-\), which gives a positive measure. On finite measure spaces with \(\abs \nu(X) < \infty\), we can define a norm \(\norm \nu = \abs\nu(X)\), so the space of measures is a normed vector space. This space is actually really nice – not only is it a Banach space, but it’s also the dual of the space of compactly supported continuous functions on \(X\), \(C_c(X)\), with the sup norm2.

lebesgue & radon & nikodym

we now move to the main event – decomposing a measure with respect to another. Here’s the statement of the Lebesgue-Radon-Nikodym theorem that we’ll now prove twice:

Theorem. (Lebesgue-Radon-Nikodym) Suppose \(\nu\) and \(\mu\) are finite positive measures on \(X\). There exist unique finite measures \(\nu_a\) and \(\nu_s\) such that \(\nu = \nu_a + \nu_s\), \(\nu_a \ll \mu\), and \(\nu_s \perp \mu\). Moreover, there is a unique \(f \in L^1(\mu)\) with \(\nu_a(E) = \int_E f \, d\mu\).

The second part of this theorem basically says something like “u-substitution works for measures,” in the sense that the \(L^1\) function \(f\) is the factor we need to “change variables” when integrating against \(\nu\) to integrating against \(\mu\). In particular, if \(\nu \ll \mu\), then \(\nu_s = \nu\) and we call the function \(f\) the “Radon-Nikodym derivative” of \(\nu\) with respect to \(\mu\), as one would compute when doing a u-substitution (except we’re doing this with measures now and not doing Riemann integration). In particular, sometimes people denote \(f\) by \(\frac{d\nu}{d\mu}\), which seems kind of unhinged if you think about it too hard.

I’m going to present two arguments for this theorem here, one out of Folland proper (in Section 3.2) and the other my solution to an exercise in Folland Chapter 6 (supplemented by a similar argument in Stein-Shakarchi, Section 6.4).

the analyst’s way

the way that feels most analytical is to just try and construct something that should end up being the Radon-Nikodym derivative, and then showing that it is actually what we want. As an analyst, though, often times showing the construction works boils down to “if it doesn’t, something goes bad” as with the excluded-middle part of our argument for Hahn decomposition above.

Here, to do this we consider the collection \(\mc F\) of all measurable functions \(f : [0, \infty]\) (with extended values) such that \(f \, d\mu \leq d\nu\), i.e. for every measurable \(A\), \(\int_A f \, d\mu \leq \nu(A)\). Observe that \(\mc F\) is non-empty, since the zero function \(f = 0\) satisfies this condition.

Note that \(\mc F\) is closed under taking maxima. For any \(f, g \in \mc F\), with \(h = \max(f, g)\) pointwise, then let \(A = \set{x : f(x) > g(x)}\). Then for any \(E\) measurable, we have that

\[\int_E h \, d\mu = \int_{E \cap A} h \, d\mu + \int_{E - A} h \, d\mu = \int_{E \cap A} f \, d\mu + \int_{E - A} g \, d\mu \leq \nu(E \cap A) + \nu(E - A) = \nu(E)\]

by construction.

Let \(a = \sup\set{\int f \, d\mu : f \in \mc F}\), and note that since \(\int f \, d\mu \leq \nu(X) < \infty\), we have that \(a < \infty\) is finite as well. Now, pick a sequence \(\set{f_n}_{n=1}^\infty \subseteq \mc F\) such that \(\int f_n \, d\mu \to a\). Construct the increasing sequence \(\set{g_n}_{n=1}^\infty\) by taking \(g_n = \max(f_1, \dots, f_n) \in \mc F\) (note that since \(\mc F\) is closed under taking maxima of two functions, by induction it is closed by taking maxima of any finite collections). Moreover, let \(f\) be a function such that \(f(x) = \sup_{n \in \ZZ^+} f_n(x) = \lim_{n \to \infty} g_n(x)\), and by the Monotone Convergence Theorem we have that \(\int f \, d\mu = \lim_{n \to \infty} \int g_n \, d\mu\). Since \(g_n \geq f_n\) for all \(n\), we have that \(\int g_n \, d\mu \geq \int f_n \, d\mu\), and so by properties of the supremum we have that \(\int f \, d\mu = \lim_{n \to \infty} \int g_n \, d\mu = a\) by construction. \(f\) must be finite almost everywhere, so up to changing \(f\) on a null set, \(f\) is real-valued.

Let \(\nu_a\) be given by \(\nu_a(E) = \int_E f \, d\mu\). Note that \(\nu_a \ll \mu\), since if \(\mu(E) = 0\) for some measurable subset, then \(\nu_a(E) = \int_E f \, d\mu = 0\) by definition of the Lebesgue integral. Then we must have \(\nu_s\) such that \(\nu_s(E) = \nu(E) - \int_E f \, d\mu\), i.e. \(d\nu_s = d\nu - f \, d\mu\). Note that \(\nu_s\) is a positive measure since \(d\nu \geq f \, d\mu\). We show that \(\nu_s\) is singular with respect to \(\mu\) to show that this decomposition works.

Suppose for the sake of contradiction that \(\nu_s\) is not singular with respect to \(\mu\). For every \(n \in \ZZ^+\), consider the finite measure \(\nu_s - \frac 1n \mu\) on \(X\), and consider a Hahn decomposition of \(X = P_n \sqcup N_n\) with respect to \(\nu_s - \frac 1n \mu\) for every \(n\). Let \(P = \bigcup_{i=1}^\infty P_i\) and \(N_i = \bigcap_{i=1}^\infty N_i\), so that \(X = P \sqcup N\). Since \(N\) is a negative set for \(\nu_s - \frac 1n \mu\) for all \(n \in \ZZ^+\), we have that \(0 \leq \nu_s(N) \leq \frac 1n\mu(N)\), so \(\nu_s(N) = 0\). Since \(\nu_s\) is not singular with respect to \(\mu\), we must then have \(\mu(P) > 0\), so \(\mu(P_n) > 0\) for some \(n\). Therefore, \(\nu_s(P_n) \geq \frac 1n \mu(P_n) > 0\).

We use this measurable set and \(n\) to derive a contradiction. Note that \(\frac 1n \chi_{P_n} \, d\mu \leq d\nu - f \, d\mu\), so \((f + \frac 1n \chi_{P_n}) d\mu \leq d\nu\). This means that \(f + \frac 1n \chi_{P_n} \in \mc F\) as above, but \(\int (f + \frac 1n \chi_{P_n}) 7, d\mu = a + \frac 1n \mu(P_n) > a\), which contradicts the fact that \(f\) was constructed to have maximal integral. Hence \(\nu_s\) must be singular with respect to \(\mu\), as desired. (Here’s that argument style again!)

Finally, to show that this is unique, we show that \(f\) is unique \(\mu\)-almost everywhere. Suppose we had some other \(f' \in \mc F\) constructed as above, giving \(d\nu_a' = f' d\mu\) and \(d\nu_s' = d\nu - f' d\mu\). Then, \(d\nu_s' - d\nu_s = (f' - f) d\mu\). Since \(\nu_s\) and \(\nu_s'\) are both singular with respect to \(\mu\), so is \(\nu_s' - \nu_s\), but since \(d\nu_s' - d\nu_s = (f' - f) \, d\mu\), \(\nu_s ' - \nu_s \ll \mu\). This shows that \(\nu_s' - \nu_s\) must be the zero measure, so \(\nu_s = \nu_s'\) and therefore \(f \, d\mu = f' \, d\mu\). This means that \(f\) and \(f'\) \(\mu\)-almost everywhere, which shows the uniqueness.

Maybe this wasn’t too interesting – we tried the one thing that could work, but we did a lot of work to force it to work. This is where this next argument comes in – we can actually somewhat magically create our derivative out of thin air, through some Hilbert space theory instead!

dual space trickery

this argument originally came from von Neumann, and is outlined in a series of exercises in Folland, Section 6.2. The following is mostly just my solution to the said exercise. (I do remember that I had to look up how to argue this last bit carefully enough to my liking, so I think this is paraphrased from another online source that I’ve unfortunately forgotten.)3

First, define a positive measure \(\lambda = \mu + \nu\). We first show that \(\vphi : L^2(\lambda) \to \RR\) given by \(\vphi(f) = \int f \, d\nu\) is a bounded linear functional on \(L^2(\lambda)\). Note that \(\abs{\int f \, d\nu} \leq \int |f| \, d\nu\). Now, since \(\lambda = \mu + \nu\), we have that \(\nu(E) \leq \lambda(E)\), and so from the definition of the Lebesgue integral on indicator functions \(\chi_E\) we get that \(\int \chi_E \, d\nu \leq \int \chi_E \, d\lambda\), so that so by linearity \(\int f \, d\nu \leq \int f \, d\lambda\) for any \(f\) non-negative and simple. Passing to suprema, for any \(f \in L^1(\lambda)\), note that \(f \in L^1(\nu)\) since \(\norm f_{1, \nu} \leq \norm f_{1, \lambda}\). Since \(\lambda\) is a finite measure on \(X\), recall that \(L^2(\lambda) \subseteq L^1(\lambda)\), where \(\norm{f}_{1, \lambda} \leq \norm{f}_{2, \lambda} \lambda(X)^{\frac 12}\) from Hölder’s inequality (or just a version of the Cauchy-Schwarz inequality). Then \(|\int f \, d\nu| \leq \norm{f}_{1, \nu} \leq \norm{f}_{1, \lambda} \leq \lambda(X)^{\frac 12} \norm f_{2,\lambda}\), so \(\vphi\) is a bounded linear functional on \(L^2(\lambda)\) (where the linearity is easy to see since the integral is linear).

Now, by the Riesz Representation Theorem, \(\vphi(f) = \int fg \, d\lambda\) for some \(g \in L^2(\lambda)\). We first show that \(0 \leq g \leq 1\) \(\lambda\)-a.e. Consider \(f = \chi_E\) for \(E\) measurable. Then, \(\vphi(f) = \int \chi_E \, d\nu = \int_E g \, d\lambda = \nu(E) \geq 0\). If \(g \geq 0\) not \(\lambda\)-almost everywhere, then there is \(F\) measurable with \(\lambda(F) > 0\) such that \(g|_F < 0\). Then \(\int_F g \, d\lambda < 0\), contradiction. Hence \(g \geq 0\) \(\lambda\)-almost everywhere.

We can rearrange our expression for \(\vphi\) in the following way – note that \(\int fg \, d\lambda = \int fg \, d\nu + \int fg \, d\mu\). Therefore, \(\int fg \, d\mu = \int f(1-g) \, d\nu\). Again, plugging in \(f = \chi_E\), we have that \(\int_E (1-g) \, d\nu = \int_E g \, d\mu\), and adding \(\int_E (1-g) \, d\mu\) to both sides, we see that \(\int_E (1-g) \, d\lambda = \int_E 1 \, d\mu = \mu(E) \geq 0\). By similar reasoning to the above, if \(g \leq 1\) not \(\lambda\)-almost everywhere, then there is \(F\) measurable with \(\lambda(F) > 0\) and \(g|_F > 1\), so \((1-g)|_F < 0\), so \(\int_F (1-g) \, d\lambda < 0\), contradiction. Hence \(0 \leq g \leq 1\) almost everywhere, so up to changing \(g\) on a set of measure \(0\), \(g\) is bounded between \(0\) and \(1\) everywhere.

We’re now ready to define our measures. Let \(A=\set{x: g(x)<1}\) and \(B= \set{x: g(x)=1}\), and set \(\nu_{a}(E)= \nu(A \cap E)\) and \(\nu_{s}(E)=\nu(B \cap E)\). We first check that \(\nu_s \perp \mu\). Observe that by construction, \(A \cup B = X\) and \(A \cap B = \nullset\), and that \(\nu_s(A) = \nu(A \cap B) = \nu(\nullset) = 0\). To see that \(\mu(B) = 0\), observe that by plugging \(\chi_B\) into the relation we had between \(\lambda\) and \(\mu\) above, we get that \(\int_B (1-g) \, d\lambda = \mu(B) = \int_B 0 \, d\lambda = 0\) by construction of \(B\). Hence \(\nu_s \perp \mu\), as desired.

All we have to do now is to check that \(\nu_a\) is actually absolutely continuous with respect to \(\mu\). We have to be a little bit careful to work in \(A\) only so that we don’t have to deal with the singularities of the derivative we want. To that end, we consider the measurable subsets \(A_n = g^{-1}([0, 1-\frac 1n])\), and note that the sequence of functions \(\chi_{E \cap A_n} \to \chi_{E \cap A}\) in an increasing fashion as \(n \to \infty\), where \(A = \bigcup_{i=1}^\infty A_n\). Then, we have that

\[\nu_a(E) = \nu(E \cap A) = \lim_{n \to \infty} \nu(E \cap A_n) = \lim_{n \to \infty} \int \chi_{E \cap A_n} \, d\nu = \lim_{n \to \infty} \int \frac{\chi_{E \cap A_n}}{1-g} \cdot (1-g) \, d\nu.\]

Note that \(\frac{\chi_{A_n \cap E}}{1-g} \in L^2(\lambda)\) for all \(n \in \ZZ^+\), since \(1-g \geq \frac 1n\) and hence \(\frac 1{1-g} \leq n\), and therefore \(\int \left(\frac{\chi_{A_n \cap E}}{1-g}\right)^2 \, d\nu \leq \int n^2 \, d\nu = n^2 \nu(X) < \infty\). Then, applying our relation between \(\nu\) and \(\mu\) above, this right-hand side becomes \(\lim_{n \to \infty} \int \frac{g}{1-g} \chi_{A_n \cap E} \, d\mu\), and by the Monotone Convergence Theorem, this is equal to \(\int \frac{g}{1-g} \chi_{A \cap E} \, d\mu = \int_E \frac g{1-g} \chi_A \, d\mu\). This calculation gives us the Radon-Nikodym derivative \(\frac{d\nu_a}{d\mu}\) explicitly where \(d\nu_a = \frac{g}{1-g} \chi_A \, d\mu\), and shows that \(\nu_a \ll \mu\), hence proving the Lebesgue-Radon-Nikodym Theorem. (The same uniqueness argument as before suffices to show that this decomposition is unique.)

coda

with this, I hereby hang up my analysis hat. although, I was supposed to have learned a bit of ergodic theory from my analysis class 2.5 years ago too, and I didn’t absorb any of that either… maybe another day :shrug:

back to skipping through fields of flowers and counting yippeeee

  1. if \(\mu\) and \(\nu\) are finite, you can write down an equivalent condition that looks an awful lot like \(\eps\)-\(\delta\) continuity, although I’m not so sure about the “absolute”ness of it. ↩

  2. this is a version of the Riesz Representation Theorem that’s called Riesz-Markov-Kakutani on Wikipedia, and I’m butchering it a little (because it really applies only to Radon measures) but whatever I think this is still cool ↩

  3. Stein and Shakarchi have an argument along the same lines, but I’m a little skeptical of how they finish up their version of the argument because it really doesn’t wrestle with the fact that the Radon-Nikodym derivative should have a \(\chi_A\) factor in it somewhere in order to get rid of the singularities… it does look quite slick and is worth reading though ↩