2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 at 7:53
  • 1
    $\begingroup$ 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 at 8:05
1
$\begingroup$

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 $A$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! That is very enlightening $\endgroup$ – Rod H Oct 27 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.