I Pointwise convergence in Lp space

[psie]

TL;DR Summary

I'm reading about $L^p$ space and there's a remark after a proposition that confuses me.

Proposition 4.6 Let $p\in[1,\infty)$ and let $(f_n)$ be a convergent sequence in $L^p(E,\mathcal A,\mu)$ with limit $f$. Then there is a subsequence $(f_{k_n})$ that converges pointwise to $f$ except on a measurable set of zero $\mu$-measure.

Remark The result also holds for $p=\infty$, but in that case there is no need to extract a subseqeuence, as the convergence in $L^\infty$ is equivalent to uniform convergence except on a set of zero measure.

Let us mention a useful by-product of Proposition 4.6. If $(f_n)$ is a convergent sequence in $L^p(E,\mathcal A,\mu)$ with limit $f$, and if we also know that $f_n(x)\to g(x)$ $\mu(\mathrm{d}x)$-a.e., then $f=g$ $\mu$ a.e.

I fail to see why the "by-product" of Proposition 4.6 is true. Isn't $$f_n\stackrel{L^p}{\to} f \text{ and } f_n(x)\to g(x) \ \mu(\mathrm{d}x)\text{-a.e.}$$ the same statement? I am confused about how to apply Proposition 4.6 to prove the "by-product" statement. Any help is greatly appreciated.

 

[psie]

Ok, I'm a bit tired. Here's what I've gathered:

If $(f_n)$ converges to $f$ in $L^p$, then by Proposition 4.6 there is a subsequence $f_{n_k}(x)\to f(x)$ pointwise for $\mu$-almost every $x$. Since we also have that $f_n(x)\to g(x)$ pointwise for $\mu$-almost every $x$, and every subsequence of a (almost everywhere) convergent sequence converges to the same limit, it's clear that $f=g$ $\mu$-a.e.