Show that if $f_n\to f_1$ uniformly and $f_n\to f_2$ in $L^p$, then $f_1=f_2$ almost everywhere.

Solution 1:

Use the following facts:

1. $f_n\overset{L^p}{\to} f$ implies $f_n\overset{\mu}{\to} f$

Indeed, we have

$$\epsilon^p \cdot \mu(\{ x \ | \ |f_n(x) - f(x)| \ge \epsilon \}) \le\int_X |f_n(x) - f(x)|^p=\|f_n-f\|^p\overset{n\to \infty}{\longrightarrow}0$$

2. $f_n\overset{\mu}{\to} f$ implies there exists a subsequence $f_{n_k}\overset{\textrm{a. e.} }{\to}f$ as $k\to \infty$.

Indeed, there exists $n_1< n_2 < \ldots < n_k < \ldots$ such that for every $k$ we have
$$\mu(\{x \in X \ |f_{n_k}(x) - f(x) | \ge \frac{1}{k} \}) < \frac{1}{2^k}$$ It is not hard to check that $$f_{n_k} \overset{\textrm{ a. e. }}{\to} f$$

More generally, this holds if $\epsilon_k\to 0$ and $\sum_k a_k < \infty$, and $n_1< n_2 < \ldots < n_k < \ldots$ such that $$\mu(\{x \in X \ |f_{n_k}(x) - f(x) | \ge \epsilon_k \}) < a_k$$