Let $f:[0,1) \rightarrow [0,1)$ be the identity function $f(x)=x$ for $x \in [0,1)$. For $n \in \mathbb{N}$, let $f_n:[0,1) \rightarrow [0,1)$ by $f_n(x)=\lfloor2^nx\rfloor/2^n$ ; in other words, for $x \in [0,1)$ take the unique integer $i \in $ {$1,2,3,...,2^n$}$ $ such that $x \in [\frac{i-1}{2^n},\frac{i}{2^n})$ and define $f_n(x)=\frac{i-1}{2^n}.$
Find the Lebesgue integral $\int_0^1f_n(x)dx$ for each $n \in \mathbb{N}$.
I am having difficulty showing this as I cannot assume if a function is Riemann integrable, then it is Lebesgue integrable and the integrals coincide. Otherwise I would just integrate it as usual.
How can this be computed with the assumption above?