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Let $p\in [1,\infty)$. Given a Riemann integrable function $f: I \to \mathbb{R}$ defined on the closed interval $I \subseteq \mathbb{R}$, show that for every $\epsilon >0$ there exists a continuous function $g:I\to \mathbb{R}$ such that

$\Vert f-g \Vert_{L^p}<\epsilon$ , where the $L^p$-Norm ist defined by $\left( \int_I\vert h(x) \vert^p\, dx\right)^{\frac{1}{p}}$ for a Riemann integrable function $h:I\to \mathbb{R}$.

I have previously shown that for every $\epsilon>0$ there exists a step-function $S$ which satisfies the inequality. From my understand step functions are only piecewise continuous, so I thought of maybe using such a step function S and then somehow manipulate it so that it becomes continuous. But I genuinly don't know how to proceed.

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  • $\begingroup$ continuous functions of compact support and steps functions are dense in $L_p$ ($1\leq p<\infty)$ an fir such functions, the Riemann and the Lebesgue integral coincide. $\endgroup$ Jun 11 '20 at 21:41
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Hint: Suppose $f$ is the step function $f=0$ on $[0,1/2],$ $f=1$ on $(1/2,1].$ Can you find a continuous function $g$ on $[0,1]$ such that $\int_0^1|f-g|<\epsilon?$

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