# Every subsequence of $\{f_{n}\}$ has a subsequence convergent almost uniformly; does this imply convergence in measure?

I am doing some problems off of Terence Tao's blog to study for a measure theory course taught based on Royden and focusing on real valued functions, and am stuck on this problem. Have tried some kinds of countable union type arguments but am not getting very far. Any help is much appreciated.

Using the Lebesgue measure, $\{f_{n}\}$ is a sequence of measurable functions, $f_{n}: E \to \mathbb{R}$. Suppose every subsequence $\{f_{n_{k}}\}$ has a subsequence $\{f_{n_{k_{i}}}\}$ which converges almost uniformly to a measurable function $f$. Then we claim $\{f_{n}\}$ converges in measure to $f$.

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If $\{f_n\}$ does not converge to $f$ in measure, then there is a $\delta>0$ such that $$\lim_{n\rightarrow\infty} \mu( \{x\in\Bbb R : |f(x)-f_n(x)|\ge \delta\}\ne 0.$$ Then, one can select an $\alpha>0$ and an increasing sequence of integers $\{n_k\}$ such that $$\mu( \{x\in\Bbb R : |f(x)-f_{n_k}(x)|\ge\delta \}>\alpha,\quad\text{for all}\ k.$$
No subsequence of $\{f_{n_k}\}$ can converge almost uniformly to $f$.