Limit of monotonic increasing functions is again montonic increasing

Question

Let $f_n:\mathbb{R}\to [0,1]$ be monotonically increasing, that is $f_{n}(x)\leq f_n(y)$ for all $x\leq y$. If $f_n\to f$ pointwise a.e., does it follow that $f$ is also monotonically increasing, so is $f(x)\leq f(y)$ for all $x\leq y$ ?

Answer 1

Yes it does a.e. (however, it is not necessarily true that it holds everywhere)

To see this, fix $x<y$ in the set where $f_n$ converges pointwise to $f$. Now suppose that $f(x)>f(y)$. In particular, let's write $f(x) = f(y) + 2\epsilon$ for some $\epsilon>0$. Now, we know from the definition of pointwise convergence that there exists an $N$ such that for all $n>N$ $|f_n(x)-f(x)|<\epsilon$, and there also exists an $M$ such that for all $m>M$ we have that $|f_m(x)-f(x)|<\epsilon$.

Now pick a $j>\max{N,M}$ Then $f_j(y) < f_j(x) + \epsilon$, and $f_j(x) > f_j(x) + \epsilon$.

This contradicts the monotonicism of $f_j(x)$

However, we cannot get a strong result to hold everywhere. To see this, we can just change a single point, e.g. if $f_n(x) = x-\frac{1}{n}$, and $f(x) = x$ for $x$ not equal to $1$, and $f(1)=0$. Then $f_n$ converges pointwise to $f$, and $f_n$ is monotonic, but $f$ is not monotonic.