# 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.

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.

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.

• Hint: Uniform convergence implies convergence almost everywhere. Convergence in $L^p$ implies existence of subsequence converging almost everywhere. Dec 13, 2021 at 23:09

Use the following facts:

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

Indeed, we have

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

1. $$f_n\overset{\mu}{\to} f$$ 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 $$\mu(\{x \in X \ |f_{n_k}(x) - f(x) | \ge \epsilon_k \}) < a_k$$