Monotonic increasing and convergence in measure

Question

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.

Answer 1

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).