converging sequence of reals
You cannot prove it, since it is false. Just take $$f_n(x)=\begin{cases}n^2x&\text{ if }x\leqslant\frac1{2n}\\n^2\left(\frac1n-x\right)&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ otherwise.}\end{cases}$$Then, for each $x\in[0,1]$, $\lim_{n\to\infty}f_n(x)=0$. But$$(\forall n\in\Bbb N):\int_0^1f_n(x)\,\mathrm dx=\frac14.$$