Let $\mu$ be a measure on $L_m$ , ($m \ge 1$) - $\sigma$ - algebra of Lebesgue-measurable sets, such that its values on compact symmetric intervals (cubes) are equal to Lebesgue measure of those cubes. Show that then $\mu = \mathcal{L}^m$.

I know we can define Lebesgue measure by symmetric cubes, but I am not sure how to use it here.

Lebesgue measure of a set is $\inf \{ \sum_{i \in \mathbb{N}} vol(I_i) \ | \ \{I_i \} − \text{a sequence of compact intervals covering a set} \} $ So what I need to know is that the measures are equal for bounded sets (but I'm not sure how that follows from your comment) and them somehow show this for all Lebesgue measurable sets...

I also know that for any bounded measurable set $M$ and $\varepsilon >0$, there are sets $K,U$ such that $K$ is compact, $U$ is open, $K \subset M \subset U$ and $ \mathcal{L}^m (U \setminus K) < \varepsilon$.

So it appears that I need to show that for any bounded set $M \in L_m : |\mu(M) - \lambda(M)| < \varepsilon$ for any $\varepsilon >0$.

Here are my latest thoughts:

If we know that the measures agree for compact symmetric intervals, then they agree for open intervals, because we can cover an open interval with an ascending sequence of compact intervals. Next, every open set is a sum of disjoint connected components. We can obtain a compact set as a limit of descending open sets of finite measure. Then similarly we show that the measures agree for $\sigma -$ compact sets and finally there is a theorem that says that every Lebesgue-measurable set can be presented as $F \cup Z$ where $F$ is $\sigma -$ compact and $\mathcal{L}^m(Z)=0$ and $\mu(A)=\mu(F)=\mathcal{L}^m(F)=\mathcal{L}^m(A)$.

Does that make any sense? Or maybe it lacks some details?

Thank you for you help.


Your Answer

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

Browse other questions tagged or ask your own question.