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Proving that convergence of norms and convergence a.e. implies strong convergence

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I have in my notes the following theorem

Theorem

$(Y,\mathcal{F},\mu)$$\sigma-$finite measure space, $p\geqslant 1$, $\{f_n\}\subset L^p(Y)$ sequence of functions, $f\in L^p(Y)$ such that $$\lim_{n\to\infty}f_n(x)=f(x)\quad\mathrm{a.e.}$$(a.e. pointwise convergence). Then,$$\lim_{n\to\infty}||f_n||_{L^p(X)}=||f||_{L^p(X)}$$implies strong convergence in $L^p(X)$.

and I want to understand the proof I was given. My notes use Vitali convergence theorem and while I have no trouble understanding and proving the requirements concerning absolute continuity, my notes skip the proof of convergence in measure, so I would appreciate to have some feedback. How do I prove that, within the hypotheses of the theorem above,$$\{f_n\}\to f\quad\text{in measure}$$ (note that the finite-measure case is trivial due to Severini-Egorov theorem, since convergence a.e. implies almost uniform convergence, which implies convergence in measure)?


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