 Solution

For the first assertion, fix $x_0\in [0,1]$ and $\epsilon >0$. By uniform convergence, choose $N$ such that $|f_N(x)-f(x)|<\epsilon /3$ for every $x$. By continuity of $f_N$ at $x_0$, there is $\delta >0$ such that $|f_N(x)-f_N(x_0)|<\epsilon /3$ whenever $|x-x_0|<\delta$. The triangle inequality then givesThus a uniform limit of continuous functions is continuous, including at the endpoints with their relative neighborhoods.

The converse implication involving integrals is false. Set $f_1=0$ and, for $n\ge 2$, use the shrinking continuous spikes with vanishing integralTheir supports are $[1/n-1/n^2,1/n+1/n^2]$. For each $x>0$ they eventually vanish at $x$, and they vanish at zero for every $n$. Hence $f_n\to 0$ pointwise, with continuous limit. Each triangular area is $1/n^2$, so ${\int }_0^1{f}_n\to 0={\int }_0^{1}0$. Yet $\sup |f_n|=1$, excluding uniform convergence.

The final assertion is also false. Take $f_n(x)=x^n$. Then the pointwise limit isThis limit is discontinuous at one but is Riemann integrable, with integral zero. Each $f_n$ is continuous and ${\int }_0^1{f}_n=1/(n+1)\to 0$. Thus convergence of the integrals to the integral of an integrable pointwise limit does not enforce its continuity.