Real Analysis Convergence question

This may be a sledgehammer and is only a partial answer for the case when $1<p<\infty$.

Here are three facts:

1) Weakly convergent sequences are norm bounded (see for instance page 255 of this ).

2) For $1<p<\infty$, norm bounded sequences that converge pointwise a.e. are weakly convergent to their pointwise limit (see, e.g, Theorem 13.44 in Hewitt and Stromberg's, Real and Abstract Analysis).

3)Weak limits are unique.

Thus, if $f_n$ converges weakly to $h$ in $L_p$ for $1<p<\infty$ and converges pointwise to $g$ a.e, it follows that $f=g$ almost everywhere.

For $p=1$ with a finite measure space:

Weak convergence of $\{f_n\}$ insures that $\{f_n\}$ is uniformly integrable. This is known as the Dunford-Pettis theorem (see, e.g., pg. 59 here or IV.8.11 in Dunford and Schwartz' Linear Operators, Part 1, General Theory). By the Vitali Convergence Theorem (see also page 163 here), then, $\{f_n\}$ converges to $g$ in $L_1$. Then, since convergence in norm to $g$ implies weak convergence to $g$, we must have $g=h$ a.e. .

I will assume $1<p<\infty$. Since we work on a finite measure space, we can use Egoroff's theorem. Fix $\phi\in L^q$ where $q$ is the conjugate exponent (i.e. $p^{-1}+q^{-1}=1$). Fix $\varepsilon>0$, and $\delta>0$ such that if $\mu(A)\leq \delta$ then $\int_A |\phi|^q\leq \varepsilon^q$. Thanks to Egoroff's theorem, we can find $A_{\delta}$ such that $f_n\to g$ uniformly on $A_{\delta}$ and $\mu(A_{\delta}^c)\leq \delta$. So \begin{align*} \left|\int_X(g-h)\phi ~d\mu\right|&\leq \limsup_n \left(\left|\int_X(g-f_n)\phi ~d\mu \right| +\left|\int_X(f_n-h)\phi ~ d\mu \right| \right)\\\ &\leq \limsup_n\left|\int_X(g-f_n)\phi ~ d\mu \right|\\\ &\leq \limsup_n\left(\left|\int_{A_{\delta}}(g-f_n)\phi ~ d\mu \right|+\left|\int_{A_{\delta}^c}(g-f_n)\phi ~d\mu \right|\right)\\\ &\leq \limsup_n\:\sup_{A_{\delta}}|g-f_n|\left(\int_X|\phi|^qd\mu\right)^{\frac 1q}+\sup_k\lVert g-f_k\rVert_{L^p}\left(\int_{A_{\delta}}|\phi|^q\right)^{\frac 1q}\\\ &\leq \varepsilon\cdot \sup_k\lVert g-f_k\rVert_{L^p} \end{align*} so $\int_X(g-h)\phi d\mu=0$ for all $\phi \in L^q$: we conclude that $g=h$ almost everywhere.