As a part of my measure theory class each student must lecture on a certain topic. Today I lectured on the Jordan decomposition and the Radon-Nikodym Theorem. Lecturing to freshman precalculus students is one thing, lecturing to your peers and professor about measure theory is another. You encourage precalc students to ask questions and be engaged with the lecture. As the teacher you try to make the lecture more of a discussion. This is something you can do because you know and understand the material. However, this can be difficult when you don’t fully understand the subject (like measure theory). The other TA’s and I like to joke about our measure theory lectures, fake it ’til you make it. We might not have a mastery understanding of the material (yet), but we have the fundamentals. From there it’s all about looking and feeling confident when delivering your lecture.

For those who are interested, these are my notes on the lecture I gave today. The Elements of Integration and Lebesgue Measure by Robert Bartle is the reference for specific theorems, corollaries, and lemmas.

Definition. Let \lambda be a charge on \mathbb{X} and let (P,N) be a Hahn decomposition for \lambda. The positive and the negative variations of \lambda are the finite measures \lambda^+, \lambda^- defined for E in \mathbb{X} by \lambda^+=\lambda(E\cap P), \lambda^-(E)=-\lambda(E\cap N). The total variation of \lambda is the measure \left|\lambda\right| defined for E in \mathbb{X} by \left|\lambda\right|(E)=\lambda^+(E)+\lambda^-(E).

Note, by lemma 8.3, \lambda^+ and \lambda^- are well defined and do not depend on the Hahn decomposition.

Theorem 8.5 Jordan Decomposition Theorem. If \lambda is a charge on \mathbb{X} then \lambda(E)=\lambda^+(E)-\lambda^-(E) for all E in \mathbb{X}. Moreover, if \lambda=\mu-\nu where \mu and \nu are finite measures on \mathbb{X} then \lambda^+(E)\leq\mu(E) and \lambda^-(E)\leq\nu(E) for all E in \mathbb{X}.

Proof. We prove the first assertion. Let E\in X and let (P,N) be a Hahn decomposition of X. Then P\cup N=X and P\cap N=\emptyset. We have,

\displaystyle \lambda(E)=\lambda\left((E\cap P)\cup(E\cap N)\right)=\lambda^+(E)-\lambda^-(E).

We prove the second assertion. Since \mu and \nu have nonnegative values,

\displaystyle \lambda^+(E)=\lambda(E\cap P)=\mu(E\cap P)-\nu(E\cap P)\leq\mu(E\cap P)\leq\mu(E).

Similarly, \lambda^-(E)\leq\nu(E). \Box

Theorem 8.6. If f is integrable, E\in\mathbb{X}, and \lambda is defined by \lambda(E)=\int_Ef\,d\mu then

\displaystyle \lambda^+(E)=\int_Ef^+\,d\mu \quad \lambda^-(E)=\int_Ef^-\,d\mu \quad \left|\lambda\right|(E)=\int_E\left|f\right|\,d\mu.

Proof. Let P=\lbrace x\in X:f(x)\geq0\rbrace and N=\lbrace x\in X:f(x)<0\rbrace. Then X=P\cup N and P\cap N=\emptyset. Let E\in\mathbb{X}. Then

\displaystyle \lambda(E\cap P)=\int_{E\cap P}f^+\,d\mu\geq0 \qquad \lambda(E\cap N)=-\int_{E\cap N}f^-\,d\mu\leq0.

Thus, (P,N) is a Hahn decomposition of X with respect to \lambda. Now, by the definition of variation we have that

\displaystyle \lambda^+(E)=\lambda(E\cap P)=\int_Ef^+\,d\mu

\displaystyle \lambda^-(E)=-\lambda(E\cap N)=\int_Ef^-\,d\mu

\displaystyle |\lambda|(E)=\lambda^+(E)+\lambda^-(E)=\int_E|f|\,d\mu.

and the theorem is established. \Box

Definition. A measure \lambda on \mathbb{X} is said to be absolutely continuous with respect to a measure \mu on \mathbb{X} if E\in\mathbb{X} and \mu(E)=0 imply that \lambda(E)=0. In this case we write \lambda\ll\mu. A charge \lambda is absolutely continuous with respect to a charge \mu provided that the total variation \left|\lambda\right| of \lambda is absolutely continuous with respect to \left|\mu\right|.

Example. Let f\in M^+ and \lambda(E)=\int f\chi_E\,d\mu. Recall, \lambda is a measure by corollary 4.9. If \mu(E)=0 for some E\in\mathbb{X} then f\chi_E=0 \mualmost everywhere. Thus,

\displaystyle \lambda(E)=\int f\chi_E\,d\mu=\int 0\,d\mu=0

which shows that \lambda is absolutely continuous with respect to \mu.

Example. Let \lambda be Lebesgue measure and \mu the counting measure on \mathbb{R}. Then \mu(E)=0 if and only if E=\emptyset. Hence, \mu(E)=0 implies that \lambda(E)=\lambda(\emptyset)=0, which shows that \lambda\ll\mu. However, if E=\lbrace x\rbrace, the singleton set, then \lambda(E)=0 but \mu(E)=1. Thus, \mu is not absolutley continuous with respect to \lambda.

Absolute continuity can also be characterized as follows.

Lemma 8.8. Let \lambda and \mu be finite measures on \mathbb{X}. Then \lambda\ll\mu if and only if for every \epsilon>0 there exists a \delta(\epsilon)>0 such that E\in\mathbb{X} and \mu(E)<\delta(\epsilon) imply that \lambda(E)<\epsilon.

Proof. If \lambda(E)<\epsilon for any \epsilon>0 then this is necessary and sufficient for \lambda(E)=0. Conversely, suppose that there exists \epsilon>0 and sets E_n\in\mathbb{X} with \mu(E_n)<\frac{1}{2^n} and \lambda(E_n)\geq\epsilon. Let F_n=\cup_{k=n}^{\infty}E_k so that \mu(F_n)<\frac{1}{2^{n+1}} and \lambda(F_n)\geq\epsilon. Since (F_n) is a decreasing sequence of measurable sets, we have that

\displaystyle \mu\left(\bigcap_{n=1}^{\infty}F_n\right)=\lim_{n\to\infty}\mu(F_n)=0 \qquad \lambda\left(\bigcap_{n=1}^{\infty}F_n\right)=\lim_{n\to\infty}\lambda(F_n)\geq\epsilon.

Hence, \mu(E)=0 does not imply that \lambda(E)=0. \Box

Recall, corollary 4.9 states that if f\in M^+ then \lambda(E)=\int_Ef\,d\mu is a measure. Conversely, when can we express a measure \lambda as an integral with respect to \mu of some function f\in M^+?

Corollary 4.11 showed that a necessary condition for this to hold is that \lambda\ll\mu. It turns out that this condition is also sufficient when \lambda and \mu are \sigma-finite. We state this result as a theorem.

Theorem 8.9 Radon-Nikodym Theorem. Let \lambda amd \mu be \sigma-finite measures on \mathbb{X} with \lambda\ll\mu. Then there exists f\in M^+ such that \lambda(E)=\int_Ef\,d\mu for E\in\mathbb{X}. Moreover, the function f is uniquley determined \mu-almost everywhere.