Limit of continuous functions is Riemann integrable

Question

Here is an analysis problem I'm stuck on:

Let $f\in C^0([0,1])$ with $f(0)=0$ and $f$ increasing and convex. Define: $$ f_n(x) = n\big[f(x)-f(x-\tfrac{1}{n})\big] $$ Show:

1. $f(1-\tfrac{1}{n})\le\int_0^1f_n(x)\ dx\le f(1)$
2. There is a $g$ with $f_n(x)\rightarrow g(x)$ almost everywhere
3. $\int_0^1 g(x)\ dx = f(1)$

I've been able to do part 1, and I believe I can do part 2. The idea is that convexity implies $f_n(x)\ge f_m(x)$ when $n\ge m$ [I call this pointwise monotonicity below], and I use that to show $\{f_n\}$ is a Cauchy sequence in $L^1([0,1])$. There is thus a $g\in L^1([0,1])$ with $f_n\rightarrow g$ almost everywhere, and the pointwise monotonicity implies this is pointwise convergence too.

Part 3 is straightforward, except for one wrinkle: I can't seem to show $g$ is Riemann integrable. It seems like the following should be generally true, but I cannot prove it:

If $\{f_n\}$ is a pointwise monotonic sequence of continuous functions, converging to $g\in L^1([0,1])$, then $g$ is continuous almost everywhere.

Is this true? Or do I need to use more about this particular sequence of my original problem?

Answer 1

Thanks to the hints of @MarkSaving, I have an answer!

The key is that for $x<y$, for large enough $N$, we have \begin{equation*} x-\tfrac{1}{N} < x < y-\tfrac{1}{N} < y \end{equation*}

And convexity of $f$ then implies that $f_N(x)\le f_N(y)$. Since $g$ is the limit of the $f_N$, this means $g(x)\le g(y)$. So $g$ is monotonically increasing, and thus is discontinuous on a set of measure zero.