Exchanging the limit and Lebesgue integral for every set in the measure space

Question

Let $(X,\mathcal{M},\mu)$ be a measure space and let a sequence of $\mathcal{M}$-measurable functions $\{f\}_k$, where $f_k: X\rightarrow [0,\infty]$. Assume that $f_k\rightarrow f$ pointwise and that $\lim_{k\rightarrow \infty}\int_X f_k \;d\mu =\int_X f\;d\mu <\infty$. Show that:

$$\lim_{k\rightarrow \infty}\int_E f_k \;d\mu =\int_E f\;d\mu\mbox{ for every }E \in \mathcal{M}.$$

Give an example that shows this may not be true if $\int_X f\; d\mu\; =\infty$.

Answer 1

We can truncate and use a dominated convergence argument. To formalize the following argument, we should rather use a continuous map $\varphi\colon\mathbb R\to [0,1]$ such that $\varphi(x)=1$ if $0\leqslant x\leqslant N$, and $0$ if $x\geqslant N+1/N$.

Fix an integer $N$ and define $g_k:=f_k\chi_{\{f_k\geqslant N\}}$, $h_k:=f_k\chi_{\{f_k\lt N\}}$, $g:=f\chi_{\{f\geqslant N\}}$ and $h:=f\chi_{\{f\lt N\}}$. We have the bound $$\left|\int_Ef_k\mathrm d\mu-\int_Ef\mathrm d\mu\right|\leqslant \left|\int_Eh_k\mathrm d\mu-\int_Eh\mathrm d\mu\right|+\int_Xg_k\mathrm d\mu+\int_Xg\mathrm d\mu= \left|\int_Eh_k\mathrm d\mu-\int_Eh\mathrm d\mu\right|\\+\int_X f_k\mathrm d\mu-\int_Xh_k\mathrm d\mu+\int_X f\mathrm d\mu-\int_X h\mathrm d\mu.$$ Using dominated convergence for the functions bounded by $N$ and the assumption on the sequence $(f_k)$, we get $$\limsup_{k\to\infty}\left|\int_Ef_k\mathrm d\mu-\int_Ef\mathrm d\mu\right|\leqslant 2\int_Xf\mathrm d\mu-2\int_X h\mathrm d\mu=2\int_Xf\chi_{\{f\geqslant N\}}\mathrm d\mu.$$

Take $f_k:=k\chi_{(0,k^{-1})}+\chi_{(1,k)}$ on the real line with Lebesgue measure to see the assumption "$f$ integrable" is necessary.

Answer 2

Another simple solution is by Fatou's Lemma and Reverse Fatou's Lemm. We have:

\begin{equation} \limsup \int_E f_n\leq \int_Ef\leq \liminf\int_Ef_n \end{equation}

Because point wise convergence of $f_n$s implies $f(x)=\limsup f_n(x)=\liminf f_n(x)$. Therefore, $\int_Ef=\lim\int_Ef_n$.