# The Unapologetic Mathematician

## Lebesgue Decomposition

As we’ve said before, singularity and absolute continuity are diametrically opposed. And so it’s not entirely surprising that if we have two totally $\sigma$-finite signed measures $\mu$ and $\nu$, then we can break $\nu$ into two uniquely-defined pieces, $\nu_c$ and $\nu_s$, so that $\nu_c\ll\mu$, $\nu_s\perp\mu$, and $\nu=\nu_c+\nu_s$. We call such a pair the “Lebesgue decomposition” of $\nu$ with respect to $\mu$.

Since a signed measure is absolutely continuous or singular with respect to $\mu$ if and only if it’s absolutely continuous or singular with respect to $\lvert\mu\rvert$, we may as well assume that $\mu$ is a measure. And since in both cases $\nu$ is absolutely continuous or singular with respect to $\mu$ if and only if $\nu^+$ and $\nu^-$ both are, we may as well assume that $\nu$ is also a measure. And, as usual, we can break our measurable space $X$ down into the disjoint union of countably many subspaces on which both $\mu$ and $\nu$ are both totally finite. We can assemble the Lebesgue decompositions on a collection such subspaces into the Lebesgue decomposition on the whole, and so we can assume that $\mu$ and $\nu$ are totally finite.

Now that we can move on to the proof itself, the core is based on our very first result about absolute continuity: $\nu\ll\mu+\nu$. Thus the Radon-Nikodym theorem tells us that there exists a function $f$ so that

$\displaystyle\nu(E)=\int\limits_Ef\,d(\mu+\nu)=\int\limits_Ef\,d\mu+\int\limits_Ef\,d\nu$

for every measurable set $E$. Since $0\leq\nu(E)\leq\mu(E)+\nu(E)$, we must have $0\leq f\leq1$ $(\mu+\nu)$-a.e., and thus $0\leq f\leq1$ $\nu$-a.e. as well. Since

Let us define $A=\{x\in X\vert f(x)=1\}$ and $B=\{x\in X\vert0\leq f(x)<1\}$. Then we calculate

$\displaystyle\nu(A)=\int\limits_A\,d\mu+\int\limits_A\,d\nu=\mu(A)+\nu(A)$

and thus (by the finiteness of $\nu$), $\mu(A)=0$. Defining $\nu_s(E)=\nu(E\cap A)$ and $\nu_c(E)=\nu(E\cap B)$, then it’s clear that $\nu_s\perp\mu$. We still must prove that $\nu_c\ll\mu$.

If $\mu(E)=0$, then we calculate

$\displaystyle\int\limits_{E\cap B}\,d\nu=\nu(E\cap B)=\int\limits_{E\cap B}f\,d\mu+\int\limits_{E\cap B}f\,d\nu=\int\limits_{E\cap B}f\,d\nu$

and, therefore

$\displaystyle\int\limits_{E\cap B}(1-f)\,d\mu=0$

But $1-f\geq0$ $\nu$-a.e., which means that we must have $\nu_c(E)=\nu(E\cap B)=0$, and thus $\nu_c\ll\mu$.

Now, suppose $\nu=\nu_s+\nu_c$ and $\nu=\bar{\nu}_s+\bar{\nu}_c$ are two Lebesgue decompositions of $\nu$ with respect to $\mu$. Then $\nu_s-\bar{\nu}_s=\bar{\nu}_c-\nu_c$. We know that both singularity and absolute continuity pass to sums, so $\nu_s-\bar{\nu}_s$ is singular with respect to $\mu$, while $\bar{\nu}_c-\nu_c$ is absolutely continuous with respect to $\mu$. But the only way for this to happen is for them both to be zero, and thus $\nu_s=\bar{\nu}_s$ and $\nu_c-\bar{\nu}_c$.

July 14, 2010