Convergence of Lebesgue integral of a function with a restricted domain

Question

Let $f:A\subset \mathbb{R}^n\to [0,+\infty[$ a measurable integrable function over $A$, $f_{k}:A\to [0,+\infty[$ a sequence of functions such that $f_{k}(x)\to f(x)$ for all $x\in A$. Suppose that $$ \int_{A}f_k(x)\; dx\to \int_{A}f(x)\; dx $$ and let $E\subset A$. I want to prove that $$ \int_{E}f_k(x)\; dx\to \int_{E}f(x)\; dx $$ but I don't know how to start with it. I was thinking in apply Lebesgue DCT but I don't know if it's a good approximation dor the problem, or if it's there a better aproximation. any help will be appreciated

Answer 1

First assume that $\int_A f <\infty$. By Scheffe's Lemma (see below) we get $\int_A|f_k-f| \to 0$ and hence $\int_E|f_k-f| \to 0$ which gives $\int_E f_k \to \int_E f$.

Now drop the condition that $\int_A f <\infty$. If $\int_E f <\infty$ we can replace $f_n$ and $f$ by $f_n \chi_E$ and $f\chi_E$ in above proof to see that $\int_E|f_n-f| \to 0$ which implies $\int_E f_n \to \int_E f$. If $\int_Ef =\infty$ then $\infty=\int_E f \leq \lim \inf \int_E f_n$ which implies that $\int_E f_n \to \infty$ as required.

Proof of Scheffe's Lemma: $\int_A (f-f_k)^{+} \to 0$ by DCT since the integrand here is dominated by $f$. Now $\int_A (f-f_k)^{-}=\int_A (f-f_k)^{+} -\int (f-f_k) \to 0-0=0$. Adding these two we get $|\int_A|f_k-f| \to 0$.

Answer 2

Let $E\subseteq A$ be a measurable set. By Fatou's lemma, we have that $$ \int_E f(x)\; dx \leq \liminf_{k\to\infty} \int_E f_k(x) \; dx $$ and $$ \int_{A\setminus E} f(x)\; dx \leq \liminf_{k\to\infty} \int_{A\setminus E} f_k(x) \; dx. $$ Assume that $$ \int_E f(x)\; dx < \liminf_{k\to\infty} \int_E f_k(x) \; dx, $$ then we have that \begin{align*} \int_A f(x) \; dx &= \int_{E} f(x)\; dx + \int_{A\setminus E} f(x)\; dx\\ & < \liminf_{k\to\infty} \int_E f_k(x) \; dx + \liminf_{k\to\infty} \int_{A\setminus E} f_k(x) \; dx\\ & \leq \liminf_{k\to\infty} \left( \int_E f_k(x) \; dx + \int_{A\setminus E} f_k(x) \; dx \right) \\ & = \liminf_{k\to\infty} \int_A f_k(x) \; dx \end{align*} which contradicts the fact that $$ \int_A f(x) \; dx = \lim_{k\to\infty} \int_A f_k(x) \; dx. $$ Thus $$ \int_E f(x)\; dx = \liminf_{k\to\infty} \int_E f_k(x) \; dx, $$ and also $$ \int_{A\setminus E} f(x)\; dx = \liminf_{k\to\infty} \int_{A\setminus E} f_k(x) \; dx. $$ Finally \begin{align*} \limsup_{k\to\infty} \int_E f_k(x) \; dx &= \limsup_{k\to\infty} \left( \int_A f_k(x) \; dx - \int_{A\setminus E} f_k(x) \; dx \right) \\ &= \lim_{k\to\infty} \int_A f_k(x) \; dx - \liminf_{k\to\infty} \int_{A\setminus E} f_k(x)\; dx\\ &= \int_A f(x) \; dx - \int_{A\setminus E} f(x)\; dx\\ &= \int_E f(x)\; dx \end{align*} and consequently $$ \lim_{k\to\infty} \int_E f_k(x)\; dx = \int_E f(x)\; dx. $$