# monotone convergence theorem for decreasing sequence.

$\{f_n\}\ge 0$ 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.

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 $n$.

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