Exercise 1.5.8 (Convergence in measure and almost uniform convergence)

Question

Let $\left(fight)_{n\in\mathbb{N}}$ be a sequence of measurable functions from $X$ to $\mathbb{C}$, and let $f:X	o\mathbb{C}$ be another measurable function. Show that the following are equivalent:
\begin{enumerate}[label=(oman*)]
\item $\left(fight)_{n\in\mathbb{N}}$ converges in measure to $f$.
\item Every subsequence $(f_{n_j})_{j\in\mathbb{N}}$ of $\left(fight)_{n\in\mathbb{N}}$ has a further subsequence $(f_{n_{j_i}})_{i\in\mathbb{N}}$ that converges almost uniformly to $f$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{itemize}
\item[$\implies)$] This direction is easy, since every subsequence  $(f_{n_j})_{j\in\mathbb{N}}$ of $\left(f\right)_{n\in\mathbb{N}}$ converges to $f$ in measure as well, and so we can apply \href{./exercise_1-5-6}{Exercise 1.5.6}.
\item[$\impliedby)$] Now suppose that every subsequence of $\left(f\right)_{n\in\mathbb{N}}$ has a further almost uniformly convergent subsequence, and assume for the sake of contradiction that $\left(f\right)_{n\in\mathbb{N}}$ does not converge to $f$ in measure, i.e., there is a $\delta>0$ such that
$$\exists \epsilon>0 \quad \forall N \in \mathbb{N} \quad \exists n_N \geq N: \quad \mu\left(\left\{x\in X: |f_{n_N}(x)-f(x)|>\delta \right\} \right) > \epsilon.$$
By axiom of choice, we can construct a subsequence $(f_{n_N})_{N\in\mathbb{N}}$ using the above mechanism. Thanks to our choice of $n_N$, this sequence, nor any of its subsequences, converge to $f$ in measure. Therefore, they do not converge to $f$ almost uniformly - a contradiction to the theorem assumption.
\end{itemize}

Exercise 1.5.8 (Convergence in measure and almost uniform convergence)

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Exercise 1.5.8 (Convergence in measure and almost uniform convergence)

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