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Problem: If $f_n \to f$ and $g_n \to g$ in $L^p$, and $g_n$ is uniformly bounded, then $f_ng_n \to fg$ in $L^p$.

An official solution I saw for this problem looked very different. Here is my solution:

Let $\| \cdot \|$ denote the $p$-norm. Suppose $M$ is such that $|g_n| \leq M$ for all $n$. We have $$\|fg - f_ng_n\| = \|fg - fg_n + fg_n - f_ng_n\| \leq \|fg-fg_n\| + \|g_n(f-f_n)\|$$ Now $\|g_n(f-f_n)\| \leq \|M(f-f_n)\| = M \cdot \|f - f_n\|$, which converges to $0$. So $\|g_n(f-f_n)\|$ converges to $0$.

Also $$|fg_n|^p \leq |M f|^p = M^p |f|^p$$ with $M^p|f|^p$ integrable. So by the Lebesgue dominated convergence theorem, the fact that $|fg_n|^p$ converges to $|fg|^p$ pointwise means that $\int|fg_n|^p$ converges to $\int |fg|^p$. Then $\|fg_n\| = \sqrt[p]{ \int |fg_n|^p}$ converges to $\|fg\| = \sqrt[p]{\int |fg|^p}$. It can be shown that this is equivalent to saying that $\|fg - fg_n\|$ converges to $0$.

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the fact that $|fg_n|^p$ converges to $|fg|^p$ pointwise

Where did you get this fact from? Convergence in $L^p$ does not imply convergence almost everywhere.

However, the proof can be fixed. Convergence in $L^p$ does imply convergence almost everywhere for a subsequence $g_{n_k}$. So, for this subsequence your argument works and gives $f_{n_k}g_{n_k}\to fg$ in $L^p$.

Moreover, the above can be applied to any subsequence of the original sequence $(g_n)$. Thus, every subsequence of $(f_ng_n)$ has a sub-subsequence converging to $fg$. It is a general topological fact that this implies the convergence of the whole sequence to $fg$. (Here this fact is proved for real valued sequences, but the same proof works in any topological space.)

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