For all Lebesgue-Integrable function $f$, there exists two upper and lower semi-continuous functions $g,h$, such that $g\le f\le h$.

functional-analysisreal-analysis

This is one exercise from functional-analysis.

The full problem is : Let $f:X\to\mathbb{R}$ a Lebesgue-integrable function. Then $\forall s>0$, there is upper-semi-continuous function $g$ and lower-semi-continuous function $h$, such that $g\le f\le h$, and $$\int_X (h(x)-g(x))m(dx)<s$$

At first, I think of simple functions. BUT, as $f$ is not bounded and $m(X)$ could be infinite, so there is no step function could do that (Counter example:$f(x) = \frac{1}{x^2}$, $x\ge 1$ there doesn't exist simple function $\phi(x) \ge f$ meet the requirement.)

And I wonder if I can find two continuous functions. However the only theorem I can think of is Lusin'theorem. But it couldn't guarantee that continuous $g\le f$ or $g\ge f$ .

So, I get stuck. Pls let me know where should I go.

Thanks!

Best Answer

This is known as the Vitali-Caratheodory theorem; a proof can be found in Rudin's "Real and Complex Analysis" (theorem 2.25, page 56).

https://en.wikipedia.org/wiki/Vitali%E2%80%93Carath%C3%A9odory_theorem

Related Question