4
$\begingroup$

I'm having trouble proving the following proposition, which is "left to the reader".

Let $f:[a,b]\to \mathbb{R}$ be a bounded function. Let us define $$s(f) := \left\{\int \phi : \phi \text{ is simple, } \phi \leq f\right\}$$ $$S(f) := \left\{\int \phi : \phi \text{ is simple, } \phi \geq f\right\}$$ and $$\underline\int f := \sup s(f)$$ $$\overline\int f := \inf S(f).$$ Show that $f$ is Lebesgue measurable iff $\underline\int f = \overline\int f$.

Supposing that $f$ is Lebesgue measurable, I thought of approximating $f$ using two sequences of simple functions converging "from above" and "from below" to $f$, and evaluate the error in the approximation, but I couldn't develop this idea any further.

|cite|improve this question
$\endgroup$
  • $\begingroup$ There are two approaches to Lebesgue integration, one classical the other modern. It would be better if you list the definition of measurability of a function under consideration. $\endgroup$ – Megadeth Nov 19 '16 at 5:53
  • $\begingroup$ It is the same definition of measurability of a function between abstract measurable spaces, i. e., $f$ is said to be (Lebesgue) measurable if the preimage of every measurable set is measurable. $\endgroup$ – Vitor Borges Nov 19 '16 at 6:03
  • $\begingroup$ I put it this way: If I were you, I would not assume knowledge when phrasing my question in order to touch as many potential readers as possible :). $\endgroup$ – Megadeth Nov 19 '16 at 6:08
2
$\begingroup$

If we assume $$\underline\int f = \overline\int f, $$ or $\inf S(f) = \sup s(f)$ we can find simple functions $\phi_k \in S(f)$ and $\psi_k\in s(f)$ such that $\phi_k - \psi_k \to 0$ as $k \to \infty$. As, by definitions, $$ \psi_k \leq f \leq \phi_k$$ for all $k$, we have a sequence $\{ \phi_k \}_k$ of simple (and hence measurable) functions converging to $f$ pointwise. That a pointwise limit of Lebesgue measurable functions is also Lebesgue measurable is a well-known result.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Note that your pointwise argument only works because of Lebesgue measurability. For Borel measurability it would be wrong. math.stackexchange.com/questions/1095711/… . So maybe its better to rephrase your last sentence. $\endgroup$ – Adam Nov 19 '16 at 11:38
  • $\begingroup$ @Adam Thanks for pointing this out. $\endgroup$ – Sayantan Nov 20 '16 at 2:38

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.