Approximation by step function

Question

In my analysis class, I have learned that for any compactly supported and Riemann integrable function $f$ (with support $[a,b]$) and for any $\epsilon>0$, there exists a step function $g$ such that $$\int_{a}^b|f(x)-g(x)|dx<\epsilon. $$ What I wonder is that, for any $\epsilon>0$, is there also a step function $h$ such that $$ \int_{a}^{b}|f(x)-h(x)|^2dx < \epsilon? $$ And is there any connection between $g$ and $h$? Any of your help will be highly appreciated :)

Answer 1

Riemann integrable implies bounded. If $|f(x)|<M$ for all $x\in[a,b]$, then a step function $h$ with $$\int_a^b|f(x)-h(x)|\,\mathrm dx<\frac\epsilon{2M} $$ and wlog also bounded by $M$ makes $$\int_a^b|f(x)-h(x)|^2\,\mathrm dx\le\int_a^b|f(x)-h(x)|\cdot 2M\,\mathrm dx <\epsilon. $$