= Solution
For the first assertion, fix $x_0\in[0,1]$ and $\varepsilon>0$. By <uniform convergence>, choose $N$ such that $|f_N(x)-f(x)|<\varepsilon/3$ for every $x$. By continuity of $f_N$ at $x_0$, there is $\delta>0$ such that $|f_N(x)-f_N(x_0)|<\varepsilon/3$ whenever $|x-x_0|<\delta$. The <triangle inequality> then gives
$$
|f(x)-f(x_0)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|<\varepsilon.
$$
Thus \b[a uniform limit of <continuous> functions is <continuous>], including at the endpoints with their relative neighborhoods.
The converse implication involving integrals is \b[false]. Set $f_1=0$ and, for $n\ge2$, use the <shrinking continuous spikes with vanishing integral>
$$
f_n(x)=\max\{1-n^2|x-1/n|,0\}.
$$
Their 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 $f_n\to0$ pointwise, with <continuous> limit. Each triangular area is $1/n^2$, so $\int_0^1f_n\to0=\int_0^10$. Yet $\sup|f_n|=1$, excluding <uniform convergence>.
The final assertion is also \b[false]. Take $f_n(x)=x^n$. Then the <pointwise limit> is
$$
f(x)=\begin{cases}0,&0\le x<1,\\1,&x=1.\end{cases}
$$
This limit is discontinuous at one but is <Riemann integrable>, with integral zero. Each $f_n$ is <continuous> and $\int_0^1f_n=1/(n+1)\to0$. Thus convergence of the integrals to the integral of an integrable <pointwise limit> does not enforce its continuity.
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