Exercise 1.5.9 (Dominated convergence in measure)

Question

Let $(X,\mathcal{B},\mu)$ be a measure space, and let $\left(fight)_{n\in\mathbb{N}}$ be a dominated sequence of measurable functions from $X$ to $\mathbb{C}$; let $f:X	o\mathbb{C}$ be another measurable function. Show that $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in $L^1$ norm if and only if $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in measure.

Elu Thingol · Accepted Answer

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ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$
Consider the sequence of sets
$$\forall k\in\mathbb{N}: \quad A_k:=\left\{x \in X: g \geq \frac{1}{k} ight\}$$
and the resulting set
$$A:=\bigcup_{k\in\mathbb{N}} A_k =\{x \in X: g
eq 0\} \qquad 	ext{and} \qquad  X\setminus A\subseteq \bigcap_{n\in\mathbb{N}}\{x\in X: f_n(x) =0\}\cap\{x\in X: f(x) =0\}.$$
We then have for each $k\in\mathbb{N}$,
$$\intx{|f_n-f|} \leq \int_{X\setminus A_k}|f_n|+|f| \mathrm d\mu + \int_{A_k}|f_n-f|\mathrm d\mu \leq 2\int_{X\setminus A_k}|g|\mathrm d\mu + \int_{A_k}|f_n-f|\mathrm d\mu.$$
Suppose for the sake of contradiction that $\lVert f_n-fVert_{L^1}$ doesn't converge to $0$. Then we can find a $\delta>0$ and (by axiom of choice) a subsequence $(f_{n_j})_{j\in\mathbb{N}}$ such that $\lVert f_{n'}-fVert_{L^1}\geq 2\delta$. We fix $k$ such that $2\int_{X\setminus A_k}|g(x)|\mathrm d\mu(x)\leq\delta$ (such a $k$ exists by the dominated convergence theorem, since $\lim_{k	o\infty}\int_{X\setminus A_k}|g|\mathrm d\mu= \int_{X\setminus A}|g|\mathrm d\mu$ ). Then 
$$\delta\leq \int_{A_k}|f_{n_j}-f|\mathrm d\mu.$$
Now, as $A_k$ has a finite measure, we can extract a subsequence $(f_{n_{j_i}})_{i\in\mathbb{N}}$ of $(f_{n_j})_{n\in\mathbb{N}}$ which converges almost everywhere on $A_k$. Applying the classical dominated convergence theorem to this sequence, we get a contradiction.

Exercise 1.5.9 (Dominated convergence in measure)

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Exercise 1.5.9 (Dominated convergence in measure)

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