Exercise 1.5.6 (Convergence in measure implies almost uniformly convergent subsequence)

Question

Let $(X,\mathcal{B},\mu)$ be a measure space, and let $\left(fight)_{n\in\mathbb{N}}$ be a sequence of functions from $X$ to $\mathbf{C}$ that converge to a limit $f:X	o\mathbb{C}$ in measure. Then there exists a subsequence $(f_{n_j})_{j=1}^{\infty}$ that converges almost uniformly to $f$.

Elu Thingol · Accepted Answer

By theorem assumption we have the convergence in measure, i.e.,
$$\forall j \in \mathbb{N}: \quad \mu\left(\left\{x \in X: |f_n(x)-f(x)|>\frac{1}{j} ight\} ight) 	o 0.$$
In particular,
$$\forall j \in \mathbb{N}: \quad \exists N_j\in\mathbb{N} \quad \forall n \geq n_j: \quad \mu\left(\left\{x \in X: |f_n(x)-f(x)|>\frac{1}{j} ight\} ight) \leq \frac{\epsilon}{2^j}.$$
At this point, using the axiom of choice, we set our exceptional set to be
$$E := \bigcup_{j=1}^{\infty}\left\{x \in X: |f_{N_j}(x)-f(x)|>\frac{1}{j} ight\} \qquad \mu(E) \leq \sum_{j=1}^\infty \mu(E_j) =  \sum_{j=1}^\infty \frac{\epsilon}{2^j} = \epsilon $$
and our subsequence to be $(f_{N_j})_{j=1}^{\infty}$.
$$\forall j\in\mathbb{N} \quad (\exists N_j\in\mathbb{N})\quad \forall N_{j'} \geq N_j \quad \forall x \in X/E: \quad |f_{N_{j'}}(x)-f(x)| \leq \frac{1}{j}$$
as desired.

Exercise 1.5.6 (Convergence in measure implies almost uniformly convergent subsequence)

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Exercise 1.5.6 (Convergence in measure implies almost uniformly convergent subsequence)

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