How do I show that if $f$ is Riemann int., then for each $\epsilon$ there exists a step function, $g(x)$, s.t. $\int |f(x)-g(x)| \ dx < \epsilon$?

Question

How do I show that if $f$ is Riemann integrable, then for each $\epsilon>0$ there exists a step function such that $\int |f(x)-g(x)| \ dx < \epsilon$?

I know that if $f$ is Riemann integrable, then the infimum of the upper Riemann sums minus the supremum of the lower Riemann sums is less than $\epsilon$ which implies that they are equal and this value is the integral $\int_a^b f(x) \ dx$.

Answer 1

Given $\epsilon>0$, choose a partition $P=\{a=x_{0}<\cdots<x_{n}=b\}$ such that $U(f,P)-L(f,P)<\epsilon$.

Let $c_{i}=\inf_{x\in[x_{i},x_{i+1}]}f(x)$, $i=0,...,n-1$. Set $g=\displaystyle\sum_{i=0}^{n-1}c_{i}\chi_{[x_{i},x_{i+1})}$, then \begin{align*} \int_{a}^{b}|f(x)-g(x)|dx&=\sum_{i=0}^{n-1}\int_{x_{i}}^{x_{i+1}}|f(x)-c_{i}|dx\\ &=\sum_{i=0}^{n-1}\int_{x_{i}}^{x_{i+1}}(f(x)-c_{i})dx\\ &\leq\sum_{i=0}^{n-1}\int_{x_{i}}^{x_{i+1}}\left(\sup_{x\in[x_{i},x_{i+1}]}f(x)-c_{i}\right)\\ &=\sum_{i=0}^{n-1}\left(\sup_{x\in[x_{i},x_{i+1}]}f(x)-\inf_{x\in[x_{i},x_{i+1}]}f(x)\right)(x_{i+1}-x_{i})\\ &=U(f,P)-L(f,P)\\ &<\epsilon. \end{align*}

Answer 2

The following argument is for the existence of a continuous function which can approximate $f$ in the $L^{1}$ sense.

I will continue the notations used in the first argument.

Let $d_{i}=\sup_{x\in[x_{i},x_{i+1}]}f(x)$, $i=0,...,n-1$. Choose $\delta>0$ small enough such that $\delta<\dfrac{1}{3}(x_{i+1}-x_{i})$ and define $L_{i}$ the spline joining the points $(x_{i+1}-\delta,c_{i})$ and $(x_{i+1}+\delta,c_{i})$, so \begin{align*} L_{i}(x)=\dfrac{c_{i+1}-c_{i}}{2\delta}(x-(x_{i+1}+\delta))+c_{i+1}, \end{align*} for $i=0,...,n-1$.

We let \begin{align*} h&=c_{0}\chi_{[x_{0},x_{1}-\delta)}+L_{0}\chi_{[x_{1}-\delta,x_{1}+\delta)}+c_{1}\chi_{[x_{1}+\delta,x_{2}-\delta)}+L_{1}\chi_{[x_{2}-\delta,x_{2}+\delta)}\\ &~~~~+\cdots+L_{n-1}\chi_{[x_{n-1}-\delta,x_{n-1}+\delta)}+c_{n-1}\chi_{[x_{n-1}+\delta,x_{n}]}. \end{align*}

Then $h$ is continuous and \begin{align*} \int_{a}^{b}|h(x)-g(x)|dx&=\sum_{i=0}^{n-1}\left(\int_{x_{i+1}-\delta}^{x_{i}}|L_{i-1}(x)-c_{i-1}|dx+\int_{x_{i+1}}^{x_{i+1}+\delta}|L_{i}(x)-c_{i+1}|dx\right)\\ &=\sum_{i=0}^{n-1}\left(\dfrac{\delta}{2}|L_{i}(x_{i+1})-c_{i}|+\dfrac{\delta}{2}|L_{i}(x_{i+1})-c_{i+1}|\right)\\ &\leq\dfrac{\delta}{2}\sum_{i=0}^{n-1}[(d_{i}-c_{i})+(d_{i+1}-c_{i+1})]\\ &\leq\delta\sum_{i=0}^{n-1}(d_{i}-c_{i})\\ &<\dfrac{1}{3}\sum_{i=0}^{n-1}(d_{i}-c_{i})(x_{i+1}-x_{i})\\ &=\dfrac{1}{3}(U(f,P)-L(f,P))\\ &<\dfrac{\epsilon}{3}, \end{align*} so \begin{align*} \int_{a}^{b}|f(x)-h(x)|dx&\leq\int_{a}^{b}|f(x)-g(x)|dx+\int_{a}^{b}|g(x)-h(x)|dx\\ &<\epsilon+\dfrac{\epsilon}{3}\\ &=\dfrac{4}{3}\epsilon. \end{align*}