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I'm trying to prove this convergence result. Could you please elaborate on how to finish the proof?

Let $(X, \mu)$ be a probability space and $f:X \to \mathbb R_{\ge 0}$ measurable. Assume there is a sequence $(B_n)$ of measurable sets such that $\mu(B_n) \to 1$. Then $$ \int_{B_n} f \mathrm d \mu \to \int_{X} f \mathrm d \mu. $$

My attempt: Notice that $(1_{B_n} f)_n$ does not necessarily converges to $f$ $\mu$-a.e. Let $f_m := f \wedge m$. Then $f_m \in L_1 (\mu)$ for all $m$ and $f_m \nearrow f$. It follows from this result that $$ \lim_n \int_{B_n} f_m \mathrm d \mu = \int_{X} f_m \mathrm d \mu \quad \forall m. $$

The proof would be complete if we could show that $$ \lim_m \lim_n \int_{B_n} f_m \mathrm d \mu = \lim_n \int_{B_n} f \mathrm d \mu. $$

Could you elaborate on how to proceed?

Update: I already proved for the case $f \in L_1 (\mu)$ here. The remaining burden is on the case $\int_{X} f \mathrm d \mu = +\infty$ :v

Analyst
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  • Can't you just split into two cases. First if $f\in L^1(X)$, then we can use dominated convergence. Otherwise, we have $\int_X f d\mu =\infty$ and we can use Fatou's lemma to conclude that $\liminf_{n\rightarrow \infty} \int_{B_n} f d\mu \geq \int_X f d\mu = \infty$. – Severin Schraven Jul 18 '22 at 02:26
  • @SeverinSchraven $(1_{B_n} f)_n$ does not necessarily converges to $f$ $\mu$-a.e. – Analyst Jul 18 '22 at 02:27
  • You are absolutely right. Let me try again. If $f\in L^1(X)$, then we can estimate $$ \left\vert \int_{B_n} f - \int_X f \right\vert \leq \left\vert \int_{B_n} (f-f_m) \right\vert + \left\vert \int_X (f-f_m) \right\vert + \left\vert \int_{B_n} f_m - \int_X f_m \right\vert \leq 2 \left\vert \int_X (f-f_m) \right\vert + m (1-\mu(B_n)). $$ We can make the first term small by taking $m$ sufficiently large and then the second term small by taking $n$ sufficiently large (I leave it to you to deal with the epsilons). – Severin Schraven Jul 18 '22 at 02:40
  • @SeverinSchraven I already proved for the case $f \in L_1 (\mu)$ here. The remaining burden is on $\int_{X} f \mathrm d \mu = +\infty$ :v – Analyst Jul 18 '22 at 02:42
  • Well, that case works the same way. We have $$ \int_{B_n} f \geq \int_{B_n} f_m = \int_X f_m - \int_{X\setminus B_n} f_m \geq \int_X f_m - m (1-\mu(B_n)). $$ Take $m$ sufficiently large, such that the first term is large and then $n$ sufficiently large such that the second term is small. – Severin Schraven Jul 18 '22 at 02:51
  • @SeverinSchraven Thank you so much! I got it. Your approach is elegant. We have $$ \int_{B_n} f \ge \int_{B_n} f_m = \int_{X} f_m - \int_{B^c_n} f_m \quad \forall m. $$ Then $$ \lim_n \int_{B_n} f \ge \int_{X} f_m - \lim_n\int_{B^c_n} f_m = \int_{X} f_m \quad \forall m. $$ Hence $$ \lim_n \int_{B_n} f \ge \lim_m \int_{X} f_m =\int_X f. $$ – Analyst Jul 18 '22 at 03:10
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    I guess you want to put the liminf (respectively the limsup for the complement) as we do not know that the limits exists. Otherwise I agree :) it is not quite as elegant as the answer of Paresseux Nguyen, but I am glad you like it. – Severin Schraven Jul 18 '22 at 04:42

2 Answers2

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For all simple function $g$ such that $0 \le g \le f$, we have: $$\int f \mathrm{d} \mu\ge \limsup_{n} \int_{B_n} f \mathrm{d} \mu \ge \liminf_{n} \int_{B_n} f \mathrm{d} \mu \ge \liminf_{n} \int_{B_n} g \mathrm{d} \mu =\int g\mathrm{d}\mu,$$ Therefore, $$\limsup_{n} \int_{B_n} f \mathrm{d} \mu = \liminf_{n} \int_{B_n} f \mathrm{d} \mu =\int f\mathrm{d}\mu.$$ Hence, the conclusion.

Paresseux Nguyen
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Step 1. Let $f:X\to[0,\infty)$ be measurable. If $\{a_n\}$ is a positive sequence with $a_n\to\infty$, then $$\int_{\{x\in X: f(x)\leq a_n\}}f\,d\mu\to \int_X f\,d\mu.\tag{1}$$ This can be proved using Fatou: $\lim_{n\to\infty} f(x)1_{\{x\in X: f(x)\leq a_n\}}(x)=f(x)$ for all $x\in X$ (note that $f(x)<\infty$ for all $x$.)

Step 2. We may assume WLOG that $\mu(B_n)>0$ for all $n$. For any $\epsilon>0$, we consider $a_n=\frac{\epsilon}{\mu(B_n^c)}$. Since \begin{align*} \int_{\{x\in X: f(x)\leq a_n\}}f\,d\mu&=\int_{B_n\cap\{x\in X: f(x)\leq a_n\}}f\,d\mu+\int_{B_n^c\cap\{x\in X: f(x)\leq a_n\}}f\,d\mu\\ &\leq\int_{B_n}f\,d\mu+a_n\mu(B_n^c)=\int_{B_n}f\,d\mu+\epsilon, \end{align*} we have $$\liminf_{n\to\infty}\int_{\{x\in X: f(x)\leq a_n\}}f\,d\mu\leq \liminf_{n\to\infty}\int_{B_n}f\,d\mu+\epsilon.$$ Since $a_n=\frac{\epsilon}{\mu(B_n^c)}\to\infty$, $(1)$ implies that for each $\epsilon>0$, $$\liminf_{n\to\infty}\int_{\{x\in X: f(x)\leq a_n\}}f\,d\mu=\int_Xf\,d\mu.$$ Hence, $$\int_Xf\,d\mu\leq \liminf_{n\to\infty}\int_{B_n}f\,d\mu+\epsilon,\qquad \forall \epsilon>0,$$ i.e., $$\int_Xf\,d\mu\leq \liminf_{n\to\infty}\int_{B_n}f\,d\mu\leq \limsup_{n\to\infty}\int_{B_n}f\,d\mu\leq \int_Xf\,d\mu.$$

Feng
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