I want to calculate $\int_0^1 x^2\, dx$ directly via the definition of the Lebesgue integral. I defined a sequence of step functions for $f(x)=x^2$ as follows: $$f_n:=\sum_{k=1}^{2^n} f\left(\frac{k-1}{2^n} \right) \chi_{[\frac{k-1}{2^n},\frac{k}{2^n}]} .$$
Now I have shown that $f_n \leq f_{n+1}$ for every $n \in \{1,2,\ldots, 2^n\}$, but I am having trouble showing that $f_n\to f$ pointwise (although, it is pictorially obvious). Let $x \in [0,1]$. Then $$ \lim_{n \to \infty} f_n(x)=\lim_{n \to \infty} \sum_{k=1}^{2^n} f\left(\frac{k-1}{2^n} \right) \chi_{[\frac{k-1}{2^n},\frac{k}{2^n}]}(x). $$ This $x$ can fall in any one of the intervals, but the sizes of the intervals are becoming smaller and smaller as $n \to \infty$, so I am not sure about what to do.
First of all, for simplicity define your functions $f_n$ with indicator functions $ \chi_{[\frac{k-1}{2^n},\frac{k}{2^n}[}$ instead, so you avoid doubling the value at each junction. You can exclude $x=1$ (measure $0$) or make it a special case.
Then let x $\in [0,1[$, $\forall n \in \mathbb{N}^*$ define $x_n=\lfloor 2^n x \rfloor / 2^n$. We have $f_n(x)=f(x_n)$. We also have $|x_n-x|<1/2^n$ and so we find that $|f_n(x)-f(x)|= |x^2-x_n^2|< 1/2^{n-1} \to_{n \to \infty} 0$.
This proves the pointwise convergence.