Let $(f_n)$ be a sequence of continues functions in $[a, b]$ that uniformly converge to $f(x)$ I need to prove that $$\lim_{n\to\infty} \int_a^b |f(x) - f_n(x)|^p dx =0$$ for every $p>0$.
Let $g_n(x)=|f(x) - f_n(x)|^p$, because $(f_n)$ are continues functions in $[a, b]$ so does $f(x)$ and as a result $g_n(x)$ too. thus $$\lim_{n\to\infty}g_n(x)=\lim_{n\to\infty}|f(x) - f_n(x)|^p=0=g(x)$$ In order to prove that $\lim_{n\to\infty} \int_a^b |f(x) - f_n(x)|^p dx =0$ for every $p>0$, I first need to prove that $g_n(x)$ uniformly converge to $g(x)$. $(f_n)$ uniformly converge to $f(x)$ hence $$\lim_{n\to\infty} sup\{|f_n(x)-f(x)|, a\le x\le b\}=0$$ so there is an $N$ that for every $n>N$ $$|f_n(x)-f(x)|\le sup\{|f_n(x)-f(x)|, a\le x\le b\}<1$$ let's assume $p\ge 1$ for every $n>N$ $$|f_n(x)-f(x)|^p\le |f_n(x)-f(x)|\Longrightarrow sup\{|f_n(x)-f(x)|^p, a\le x\le b\} \le sup\{|f_n(x)-f(x)|, a\le x\le b\}$$ hence $$\lim_{n\to\infty} sup\{|f_n(x)-f(x)|^p, a\le x\le b\}=0$$ and $g_n(x)$ uniformly converge to $g(x)$. but this is not the case for $0<p<1$, and I haven't found a way to show it does uniformly converge for said p.
Let $p\in(0,\infty)$. Fix $\varepsilon>0$. There exists a $N\in\mathbb N$ such that $\sup_{[a,b]}|f(x)-f_n(x)|<\varepsilon^{1/p}$ for all $n>N$. Thus for any $x\in [a,b]$ and $n>N$ we have $|f(x)-f_n(x)|^p<\varepsilon$. As this is true for any $x$ independent of $N$ we have $|f(x)-f_n(x)|^p\to 0$ uniformly.