# A general relation between two Borel measures.

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.

• 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. – Rod H Oct 27 '20 at 7:53
• Lebesgue decomposoition can be proved using Riesz Theoerm for Hilbert spaces. That may be why this appeared in a FA course. – 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 $$A$$.