For a function $f \in L^2[0,T]$, and a uniform partition $P = \{0=t_0, t_1, \ldots, t_n = T\}$ of the domain, we can define a step function approximation as the average value over each interval in the partition, $$f_n = \sum_{k=1}^n \frac nT \chi_{[t_{k-1}, t_k]} \int_{t_{k-1}}^{t_k} f(t)dt.$$ I'm trying to prove that as $n\to \infty$ and the partition becomes arbitrarily fine, $f_n \to f$ in $L^2$. So far, I've got \begin{align*} \|f - f_n\|_2^2 & = \sum_{k=1}^n \int_{t_{k-1}}^{t_k} \left( f(t) - \frac nT \int_{t_{k-1}}^{t_k} f(x) dx \right )^2 dt \\ & =\sum_{k=1}^n \int_{t_{k-1}}^{t_k} \left(\frac nT \int_{t_{k-1}}^{t_k} f(t) dx - \frac nT \int_{t_{k-1}}^{t_k} f(x) dx \right )^2dt \\ & =\sum_{k=1}^n \int_{t_{k-1}}^{t_k} \left(\frac nT \int_{t_{k-1}}^{t_k} f(t) - f(x) dx \right )^2dt\\ & \leq \sum_{k=1}^n \frac nT \int_{t_{k-1}}^{t_k} \int_{t_{k-1}}^{t_k}(f(t) - f(x))^2dxdt \end{align*} Now, I know I can make the inner integral less than any positive $\varepsilon$ with $n\to \infty$ via absolute continuity of the integral. Evaluating the outer integral then gives a $\frac Tn$, so I'm left with $\sum_{k=1}^n \varepsilon$, which goes to infinity in the limit. So I clearly need to capture some other source of "smallness" within my integral.
Since $t$ and $x$ both lie within the interval $[t_{k-1}, t_k]$, I know that they are no more than $\frac nT$ apart. However, my attempts to exploit this with a change of variables have been fruitless. Is there something I'm missing here?
Addendum: I know I can prove convergence of $f_n \to f$ by showing that $\| f_n \|_2$ is uniformly bounded by $\|f\|_2$, and proving convergence for continuous f. However, I'm curious as to whether there's a way to use absolute continuity of the inner integral and the nearness of $t$ and $x$.