Suppose $\lambda,\mu$ are two $\sigma$-finite Borel measures (that is, measures defined on the Borel algebra in $\mathbb{R}$) where $\lambda(\mathbb{R})<+\infty$. Prove that $\exists B,\mu(B)=0\text{ and }f\in L^1(\mathbb{R},\mu)\ s.t.\int_Af\mathrm{d}\mu=\lambda(A\backslash B)$.

Here is my attempts: First thing to notice here is $\forall A\cap B=\varnothing, \int_{A}f\mathrm{d}\mu=\lambda(A)$, which means $B$ has to contain the 'biggest' measure-zero set (for example, let $\lambda(\{0\})\neq 0=\mu(\{0\})$, then $\{0\}\subset B$, otherwise $\int_{\{0\}}f\mathrm{d} \mu=0$), but something like Zorn's Lemma clearly doesn't apply here.

Then I think $\lambda,\mu$ could be considered as two Lebesgue-Stieltjes measures (induced by right continuous functions $ f(x)=\mu(0,x])$ and $g(x)=\lambda(0,x]$, for instance) by $\sigma$-finiteness, and $\int_{A}g\circ f^{-1}\mathrm{d}\mu=\lambda({A})$ might stand. So I need to find a $B$ such that $f^{-1}$ is well defined on $X-B$. Since $f$ is monotone, I think $B$ might have something to do with the set 'Int$\{x:\frac{\mathrm{d}f}{\mathrm{d}x} $ exists and equals to $0\}$', which is another representation of my first idea, but I don't know how to choose this $B$ properly.

Now I think I might have complicated the problem...Any help or hint would be appreciated.

  • $\begingroup$ By the way, this is a problem I encountered during my functional-analysis course, but I don't quite get its correlation with functional analysis. $\endgroup$ – Rod H Oct 27 '20 at 7:53
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    $\begingroup$ Lebesgue decomposoition can be proved using Riesz Theoerm for Hilbert spaces. That may be why this appeared in a FA course. $\endgroup$ – Kavi Rama Murthy Oct 27 '20 at 8:05

This is an exercise on Lebesgue decomposition. We can write $\lambda$ as $\lambda_1+\lambda_2$ where $\lambda_1 << \mu$ and $\lambda_2 \perp \mu$. There exists $B$ such that $\mu (B)=0$ and $\lambda_2 (B^{c})=0$. Also there exists $f \in L^{1}(\mu)$ such that $\lambda_1 (E) =\int _E fd\mu$ for all $E$. Now $\lambda (A\setminus B)=\int_{A\setminus B} fd\mu+\lambda_2 (A\setminus B)=\int_A fd\mu$ for all $A$.

  • $\begingroup$ Thank you! That is very enlightening $\endgroup$ – Rod H Oct 27 '20 at 8:10

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