0
$\begingroup$

Assume that you have a measurespace $(A,\mathcal{A},\mu)$. And you sequence of measurable functions $f_n \rightarrow \mathbb{R}$, that are increasing, and each function is bounded below by a common value $-M$.

Do we then have that $\lim\limits_{n \rightarrow \infty}\int\limits_{A}f_n(x)d\mu=\int\limits_{A}\lim\limits_{n \rightarrow \infty}f_nn(x)d\mu$?

I am able to prove this for a finite measure space by considering the non-negative and increasing sequence $\{f_n+M\}$ and using the monotone convergint theorem. But does it hold for a measure-space with infinite measure?

The reason I don't get it to work with a general measure space is that the integral of the constant function $M$ may not be finite, so I get in a situation where I can't cancel the parts.

$\endgroup$
2
$\begingroup$

Not true. On the real line with Lebesgue measure let $f_n(x)=-1$ for $x \geq n$ and $0$ for $x <n$. Then $f_n \geq -1, (f_n)$ is increasing and $\lim \int f_n=-\infty \neq 0 =\int \lim f_n$

$\endgroup$
1
  • 1
    $\begingroup$ Thank you very much! $\endgroup$
    – user394334
    Jul 18 '20 at 0:15

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.