If for each $n\in\mathbb{N}$, $f_n$ is monotonic increasing on [0,1] and $f_n\rightarrow f$ in measure, then $f_n\rightarrow f$ at every x at which f is continuous. I'm not sure whether this is right or not. Also I don't have any idea to deal with it.
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If $x$ is allowed to be either of the endpoints, it is not true: take $f=1$, $f_n(x)=x^{1/n}$, this will not converge at $0$.
For interior points it is true. Suppose $\exists x, \epsilon$ s.t. $f$ is continuous at $x$ and $\forall n$ $|f_n(x)-f(x)|>\epsilon$. Take $\delta$ s.t. $|f(y)-f(x)|\leq \epsilon/2$ for all $y$ s.t. $|y-x|\leq\delta$, and take $N$ s.t. $$\lambda(z:|f_N(z)-f(z)|>\epsilon/2)<\delta/2.\tag{1}$$ Now either $f_N(x)>f(x)+\epsilon$ or $f_N(x)<f(x)-\epsilon$. In the first case notice that for all $y\in[x,x+\delta]$, $$f_N(y)\geq f_N(x)>f(x)+\epsilon\geq f(y)+\epsilon/2.$$ contradicting (1). The other case is symmetric (this is where we use that $x$ is in the interior).
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$\begingroup$ But if $f_n(x)$ fails to converge to $f(x)$, then you only have the existence of $\epsilon>0$ and a subsequence $n(k)$ such that $|f_{n(k)}(x)-f(x)|>\epsilon$ for all $k$. $\endgroup$ Commented Apr 26, 2016 at 0:36
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$\begingroup$ @John Dawkins: You're right, I was sloppy. Nevertheless (1) holds for all large enough $N$, so the arguments doesn't really change. $\endgroup$– m7eCommented Apr 26, 2016 at 7:52