The paragraph is given in the picture below:
**My question is: **
1- Why for a sequence $\{f_{n}\}$ and a function $f$ in $C[a,b],$ $f_{n} \rightarrow f$ in $C[a,b],$ normed by the maximum norm, iff $f_{n} \rightarrow f$ uniformly on $[a,b].$ could anyone give me a proof for such statement please?
2-Also, I do not understand why the second statement is true, could anyone explain it for me, please?
If $f_n$ and $f$ are continuous and $\|f_n-f\|_{\infty} \to 0$ and $\epsilon >0$ then there exists $n_0$ such that $n >n_0$ implies $|f_n(x)-f(x)| \leq \epsilon$ almost everywhere. If $E$ is any set of measure $0$ in $[a,b]$ the $E^{c}$ is dense. Hence $|f_n(x)-f(x)| \leq \epsilon$ holds on a dense set of points and continuity shows that the inequality holds for all $x$. Hence $f_n \to f$ uniformly.
Suppose $\|f_n-f\|_{\infty} \to 0$ but we do not have any continuity assumption. Let $k$ be a postive integer Then there exists $n_0$ such that $n >n_0$ implies $|f_n(x)-f(x)| \leq \frac 1 k$ almost everywhere. Let $E_k$ be set of measure $0$ such that the inequality holds whenever $x \notin E_k$. Let $E=\bigcup_k E_k$. Then the inequality holds for every $k$ for $x \notin E$. Hence $f_n \to f$ uniformly on the complement of $E$.