1
$\begingroup$

I'm trying to solve the following problem (self-study) regarding Lebesgue measure in $\mathbb{R}$:

Let $E \subset [a,b]$, and suppose there is a continuous function $f:[a,b]\to\mathbb{R}$ such that for any subinterval $I \subset [a,b]$, $\mu(E \cap I) = \int_I f \, d\mu$. Show that either $\mu(E) = 0$ or $\mu(E) = b - a$.

I don't seem to be getting much of anywhere, so a gentle nudge in the right direction would be appreciated. It did occur to me that the continuity of $f$ must be important, so I used the "integral mean value theorem" to show that for all $I_k = (a_k, b_k)$, there is a $c_k$ such that $\mu(E\cup I_k) = \int_{I_k} f = f(c_k)(b_k - a_k)$. Still, I'm not sure that's getting me anywhere...

$\endgroup$
  • 1
    $\begingroup$ Sorry, 'm' was a typo and $\mu$ is meant to refer to Lebesgue measure. $\endgroup$ – bosmacs Sep 3 '13 at 20:17
1
$\begingroup$

Notice that $0\leqslant f\leqslant 1$.

As an open set is a countable disjoint union of open intervals, we have $\mu(E\cap O)=\int_Of\mathrm d\mu$ for each open subset of $[a,b]$, and by regularity, this can be extended to every Borel subset. We thus get that for any Borel subset $S$ of $[a,b]$, $$\int_S(\chi_E(x)-f(x))\mathrm dx=0,$$ hence $f=\chi_E$ almost everywhere. Conclude, using continuity of $f$.

$\endgroup$
  • $\begingroup$ What was $S$? An arbitrary Borel subset of $[a,b]$? $\endgroup$ – bosmacs Sep 3 '13 at 20:39
  • $\begingroup$ Yes. Edited now. $\endgroup$ – Davide Giraudo Sep 3 '13 at 20:40

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.