Show that for every set $A \subset \mathbb R^n$ lebesgue measurable $\int_{A} f_n dx\rightarrow \int_{A} f dx.$

Question

Let $f_n$ and $f$ functions of $L_1\mathbb R^n$ such that $\lim_{n \rightarrow \infty} f_n = f$ almost always in $\mathbb R^n$ and $\int_{\mathbb R^n} |f_n|\ dx\rightarrow \int_{\mathbb R^n} |f|\ dx$. Show that for every set $A \subset \mathbb R^n$ lebesgue measurable $$\int_{A} f_n\ dx\rightarrow \int_{A} f\ dx.$$ How can I solve this problem?

Answer 1

Here is a hint to get you started: Apply Fatou's lemma to $$ |f_n|+|f|-|f_n-f|\geq 0 $$ to find $f_n\to f$ in $L^1$.

Answer 2

Note that \begin{eqnarray*} & & |f_{n}|\\ & = & |\left(f_{n}-f\right)+f|\\ & \leq & \left|f_{n}-f\right|+\left|f\right|. \end{eqnarray*} Therefore, $|f_{n}|-|f|-|f_{n}-f|\leq0\leq2|f|$. On the other hand, \begin{eqnarray*} & & \left|f_{n}-f\right|\\ & \leq & |f_{n}|+|f|, \end{eqnarray*} re-arranging terms yields $|f_{n}|-|f|-|f_{n}-f|\geq-2|f|$. It follows that $$\left||f_{n}|-|f|-|f_{n}-f|\right|\leq2|f|.$$

Observe that $|f_{n}|-|f|-|f_{n}-f|\rightarrow0$ pointwisely a.e.. By Lebesgue Dominated Convergence Theorem, we have that $$ \int\left(|f_{n}|-|f|-|f_{n}-f|\right)dm\rightarrow0. $$ It is given that $\int\left(|f_{n}|-|f|\right)dm\rightarrow0$, so $\int|f_{n}-f|dm\rightarrow0$. In particular, for any measurable set $A$, $\left|\int_{A}f_{n}dm-\int_{A}fdm\right|\leq\int_{A}|f_{n}-f|dm\leq\int|f_{n}-f|dm\rightarrow0$.

Answer 3

This is an easy application of a "generalized" dominated convergence theorem. You have $|f_n\chi_{A}| \leq |f_n|$, and $\int |f_n| \to \int |f|$. Since $f_n\chi_{A} \to f\chi_{A}$ almost everywhere, the generalized dominated convergence theorem gives $\int f_n\chi_{A} \to \int f\chi_{A}$