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 $f\chi_E=0$ $\mu$almost 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.