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Theorem (Lebesgue Decomposition Theorem)

Let $\lambda, \mu$ be $\sigma$ -finite measures on $\mathbb{X}$. Then there exists measures $\lambda_{1}, \lambda_{2}$ such that $\lambda_{1} \perp \mu, \lambda_{2} \ll \mu$ and $\lambda=\lambda_{1}+\lambda_{2} .$ The measures $\lambda_{1}, \lambda_{2}$ are unique.

Proof. Let $\nu=\lambda+\mu .$ Thus $\nu$ is $\sigma$-finite and both $\lambda$ and $\mu$ are absolutely continuous with respect to $\nu .$ We can now use the Radon Nikodym Theorem to see there exists $f, g \in M^{+}$ such that for all $A \in \mathbb{X}$ $\mu(A)=\int_{A} f \mathrm{d} \nu \text { and } \lambda(A)=\int_{A} g \mathrm{d} \nu.$

Take the set $B=\{x: f(x)=0\}$ and note that $\mu(B)=0 .$ We define $\lambda_{1}$ and $\lambda_{2}$ to be the measures such that $\lambda_{1}(A)=\lambda(A \cap B) \text { and } \lambda_{2}(A)=\lambda\left(A \cap B^{c}\right).$ $\lambda_{1}(X \backslash B)=0$ so we can see it is singular with respect to $\mu .$

To show $\lambda_{2}$ is absolutely continuous with respect to $\mu$ we let $A \in \mathbb{X}$ satisfy $\mu(A)=0 .$ We then have that $f(x)=0$ for $\nu$ almost all $x \in A$ and thus for $\lambda$ almost all $x$.

Therefore $\lambda_{2}(A)=\lambda(A \cap B^c)=0$ and so $\lambda_{2}$ is absolutely continuous with respect to $\mu$.

How is the last sentence obtained? $\lambda_{2}(A)=\lambda(A \cap B^c)=0$? If $f=0$ $\lambda$-almost everywhere in $A$, then only a subset of $A$, say $A'$, satisfy $\lambda (A')=0$. How do we now that $A \cap B^c \subseteq A'$? Thanks

Loli
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  • Because $B$ is the set where $f$ is zero. – Norse Dec 31 '19 at 15:55
  • @Norse I thought of that but, it's not clear to me why this is the case, because $\lambda$ is not defined with respect to $f$, but wrt to $g$. How does the definition of $B$ imply the conclusion in that case? Thanks – Loli Dec 31 '19 at 15:58
  • Are you completely sure it should be $\lambda_1$?. In the second line it says that $\lambda_2$ is absolutely cont. wrt. $\mu$ not $\lambda_1$ – Norse Dec 31 '19 at 16:00
  • @Norse Yes, that's a typo. It should be $\lambda_2$ ! – Loli Dec 31 '19 at 16:01
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    Then it should read $\lambda_2(A)=\lambda(A\cap B^c)$ – Ciarán Ó Raghaillaigh Dec 31 '19 at 16:09
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    @CiaránÓRaghaillaigh Yes, I understand it now. It makes sense now, the typo is from my text book... so I didn't see it until now. but it answers the question. It's not too hard to deduce the conclusion once the typo corrected – Loli Dec 31 '19 at 16:13
  • @CiaránÓRaghaillaigh can you please see if my answer under is correct please? thanks – Loli Dec 31 '19 at 16:28
  • @Norse can you please see if my answer under is correct please? thanks – Loli Dec 31 '19 at 16:28

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To answer my own question after correction of the typo (by Norse, thanks again).

We have $\lambda_2(A)=\lambda(A\cap B^c)$, $\lambda(A')=0$ and $A'\subseteq A$.
So $A'\subseteq B^c $ (because $f(A')\neq0$) and $ A' \subseteq A $, thus $A' \subseteq A\cap B^c$.

Also because $f(B^c)\neq0$, it follows $f(A\cap B^c) \neq 0$ so because $A\cap B^c \ \subseteq A$ we must have $A\cap B^c \ \subseteq A'$, therefore $ A\cap B^c = A'$, and we are finish.

Is this correct?

Loli
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