Continuous limit of continuous functions with a distance condition Let $\{f_n\}_n$ be a sequence of continuous functions $f_n\colon[0,1]\to\Bbb R$ and let $U_n=\{\,x\in[0,1]\mid f_n(x)>1\,\}$.
We know that $\forall x\in[0,1]\colon \lim_{n\to\infty}f_n(x)=0$ does not imply that $f_n\to 0$ uniformly. However, the usual counterexamples all have $\lim_{n\to\infty}\mu(U_n)=0$. Is there a counterexample where $\mu(U_n)\ge p>0$ for all $n$? And if so, for which $p$ (certainly not for $p=1$)?
Intuitively, my answer is "no", because it feels as if for "random" $x\in[0,1]$, the "probability" that $f_n(x)\ge1$ should be at least $p$, contradicting $f_n(x)\to 0$.
 A: We must have $\mu(U_n)\to 0$. It also doesn't matter that the $f_n$'s are continuous or that they converge pointwise everywhere, mere measurability and pointwise a.e. convergence is fine.
Proof. Let $\epsilon > 0$ be given. Since $f_n\to 0$ pointwise a.e. on the finite measure space $[0,1]$, Egorov's theorem tells us that $f_n \to 0$ almost uniformly in the sense that there is a measurable set $A_\epsilon\subset [0,1]$ such that $\mu\big([0,1]\smallsetminus A_\epsilon\big) < \epsilon$ and $f_n\to 0$ uniformly on $A_\epsilon$. Hence for all sufficiently large $n$, we have $\mu\big(\{|f_n| > 1\}\big) \le \mu\big([0,1]\smallsetminus A_\epsilon\big) < \epsilon$. Thus $\mu(U_n)\to 0$.
Note that there was nothing special about $1$. We could have replaced it with any $\alpha > 0$.
A: In fact it can be generalisable, for $\epsilon > 0$ let $U_n(\epsilon) = \{x \in [0,1] : f_n (x) \ge \epsilon\}$ and let $X$ a random variable uniformly distributed on $[0,1]$. Since $f_n \to 0$ pointwise then $f_n(X) \to 0 \text{ a.s.}$ then $f_n(X) \to 0$ in probability. and so $\mu (U_n(\epsilon))=\mathbb P(f_n (X) \ge \epsilon ) \to 0$
