0
$\begingroup$

$\{f_n\}\ge 0$ is a decreasing sequence in $(X,\mathcal F,\mu)$, and $\mu(X)<\infty$, I want to prove: $$\int_X \lim_{n\to\infty} f_n\,d\mu=\lim_{n\to\infty}\int_X f_n\,d\mu$$

there is an counterexample for $\mu(X)=\infty$: $f_n(x)=1_{[n,\infty)}(x)$, but I didn't find the counterexample for $\mu(X)<\infty$

from the monotone convergence theorem for increasing sequence, $\{f_1-f_n\}\ge 0$ is increasing, so we have $$\int_X \lim_{n\to\infty}(f_1-f_n)\,d\mu=\lim_{n\to\infty}\int_X (f_1-f_n)\,d\mu$$

that is $$\int_X f_1\,d\mu-\int_X \lim_{n\to\infty}f_n\,d\mu=\int_X f_1\,d\mu-\lim_{n\to\infty}\int_X f_n,d\mu$$

if $\int_X f_1 \,d\mu=\infty$, we can't cancel it in the equation above.

Do we have an example that: $\forall n, \int_X f_n\,d\mu=\infty$, but $\int_X\lim f_n\,d\mu<\infty$? where $\mu(X)<\infty$ and $f_n$ are non-negative decreasing sequence.

$\endgroup$

1 Answer 1

3
$\begingroup$

Let $f_1$ be a positive function with $\int f_1d\mu=+\infty$.

This can be achieved under the extra condition $\mu(X)<\infty$.

Let $(c_n)_n$ be a decreasing sequence with $c_n>0$, $c_1=1$ and $\lim_{n\to\infty}c_n=0$.

Let $f_n:=c_nf_1$.

Then $(f_n)_n$ is a sequence of positive decreasing functions.

This with $\lim_{n\to\infty} f_n(x)=0$ for each $x$ so that $\int\lim f_nd\mu=0$.

Next to that we have $\int f_nd\mu=c_n\int f_1d\mu=+\infty$ for each $n$.

$\endgroup$
1
  • 2
    $\begingroup$ Exactly. We could use $f_1= 1/x$ on $(0,1)$ with the Lebesgue measure, $c_n=1/n$ $\endgroup$
    – Dosidis
    Apr 8, 2016 at 13:58

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.