We have that statement that if $\{f_n\}$ are positive real-valued measurable functions $X \rightarrow \mathbb{R}$ such that $f_n$ converges pointwise to $f$ everywhere on $X$ and $\lim \int f_n = \int f < \infty$, then $\lim \int_E f_n = \int_E f$ for every measurable $E$. There is a solution given here:
Prove that $\int_E fd\mu = \lim \int_E f_n d\mu$ for all measurable set $E$
My question is, why does this fail if $\int f = \infty$? Specifically, what is a sequence of measurable functions $f_n$ which converge pointwise to $f$ with $\lim \int f_n = \int f = \infty$, and a measurable function $E$, such that $\lim \int_E f_n < \int_E f$?
I know that the other direction must always hold, since we do not use the finiteness condition to prove that inequality. However, I cannot find any examples where the strict inequality in the other direction holds.
Best Answer
For $X=[0,+\infty)$
take $f_n(x)=n^2x^n$ if $x \in [0,1)$ and $f_n(x)=1$ on $[1+\infty)$
For $f(x)=1_{[1,+\infty)}$ we have that $+\infty=\int f_n \to +\infty=\int f$ and also $f_n \to f$ (since $x<1$)
But on $[0,1)$ we have that $\int_0^1 f_n$ do not converge to $\int_0^1 f$