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.

Question

Let $(X,\Sigma _X,\mu )$ be a measurable space, $(f_n)_{n\in\mathbb{N}}$ a sequence of $L^p(X)$ with $p\in [1,\infty ]$ and $f_1,f_2:X\to \mathbb{R}$ two measurable functions. 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.

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I proved that $|f_1(x)-f_2(x)|\leq \limsup _{n\to\infty }|f_n(x)-f_2(x)|$ for all $x\in X$.

I also know that there's an increasing sequence $(k_n)_{n\in\mathbb{N}}$ of $\mathbb{N}$ such that $\lim_{n\to\infty }|f_{k_n}(x)-f_2(x)|=\limsup_{n\to\infty}|f_n(x)-f_2(x)|$ for all $x\in X$.

I was able to prove what I asked if we assume that $|f_{k_n}(x)-f_2(x)|\leq \limsup_{n\to\infty}|f_n(x)-f_2(x)|$ for all $n\in\mathbb{N}$ and $x\in X$. However I don't know how to prove the last inequality.

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Thank you for your attention!

Answer 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 $$\mu(\{x \in X \ |f_{n_k}(x) - f(x) | \ge \epsilon_k \}) < a_k$$