Let $f_n(x)=\left\{\begin{array}{l} x^{-1/2},\ \frac{1}{n}\leq x\leq 1\\ 0,\ 0\leq x<\frac{1}{n} \end{array}\right.$
Verify $\lim_{n\rightarrow \infty}f_n(x)=x^{-1/2}$ almost always in $[0,1]$ and $\lim_{n\rightarrow\infty}\int^1_0f_n(x)=\int^1_0 \frac{dx}{x^{1/2}}$
Choose $x\in (0,1]$. Then there exists an $N\in \mathbb{N}$ such that $\frac{1}{n}<x\leq 1$ for all $n\geq N$. Hence $\lim_{n\rightarrow \infty}f_n(x)=x^{-\frac{1}{2}}$. Thus for all $x\in (0,1]$ we have that $\lim_{n\rightarrow \infty}f_n(x)=x^{-\frac{1}{2}}$. Since $\mu((0,1])=1$, the measure of the complement is zero, hence the identity holds almost everywhere. In fact notice that the identity only fails in $x=0$.
Also, there's an error in your question. It should be $\lim_{n\rightarrow \infty}\int_{0}^1f_n(x)\mathrm{d}x= \int_{0}^1\frac{1}{\sqrt{x}}\mathrm{d}x$.
EDIT: Anyhow, to show the latter, either use the monotone convergence theorem to swap the limit and integral, or explicitly calculate the limit $\lim_{n\rightarrow\infty} \int_{0}^1f_n(x)\mathrm{d}x=\lim_{n\rightarrow \infty}\int_{\frac{1}{n}}^1\frac{1}{\sqrt{x}}\mathrm{d}x$.