Consider $f_n(x) = x^n$ on the interval $[0, 1]$.

This converges pointwise to $$f = \begin{cases} 0, & \mbox{if } 0 \le x < 1\ \\ 1, & \mbox{if } x = 1 \end{cases}$$

Now I know $f_n(x)$ doesn't converge uniformly to $f$ but does it converge almost uniformly?

For any $\delta > 0$ we can can take the subset $(1 - \epsilon, 1]$ of $[0, 1]$ such that $\epsilon < \delta$. Then as the measure of this subset is $\epsilon < \delta$ and $f_n(x)$ converges to $f$ on the complement of this subset we have almost uniform convergence.

Is my understanding correct here?


You have the right idea! The sequence $f_n$ will converge almost uniformly to $f$ when: Given any $\epsilon > 0$ there exists some measurable subset $F_{\epsilon} \subset [0,1]$ such that $\mu(F_{\epsilon}) \leq \epsilon$ and $f_n \to f$ uniformly on $[0,1]\backslash F_{\epsilon}$. Note here I'm using $\mu$ as Lebesgue measure. So, take $\epsilon > 0$ (and $<1$ so we have something to talk about) and let $F_{\epsilon} = (1-\epsilon, 1]$ as you suggested. Now, $\mu(F_{\epsilon}) = \epsilon$ and so it remains to show that $f_n \to f$ uniformly on $[0,1-\epsilon]$. To do this, notice that $\sup_{0\leq x \leq 1-\epsilon}|f_n(x) - f(x)| = \sup_{0\leq x \leq 1- \epsilon}|f_n(x)| \leq (1-\epsilon)^n$, which since $\epsilon \in(0,1)$, means that $\lim_{n \to \infty} \sup_{0\leq x \leq \epsilon}|f_n(x) - f(x)| = 0$, proving uniform convergence on $[0,1]\backslash F_{\epsilon}$.


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