Showing continuous functions are dense in $L^p([0,1],\lambda)$, where $\lambda$ is Lebesgue measure

Question

Show that if $f$ is $p$-integrable then, for each $\epsilon>0$, exists a function $h$ which is continuous in $[0,1]$ s.t. $\|f-h\|_p\leq\epsilon$.

Is there any simpler way to show it than showing the fact for simple functions, then for step functions and just then showing it for continuous functions?

Answer 1

Since $f$ is $L_p$-integrable, $f$ is $L_1$-integrable. By Lusin's theorem, there exists a function $h$ which is continuous in $[0,1]$ such that $$ \mu(\{x:|f(x)-h(x)|>\epsilon^{1/p}/(2\mu(\Omega))\})<\delta $$ Let $A=\{x:|f(x)-h(x)|>\epsilon^{1/p}/(2\mu(\Omega))\}$. Then $$ \int_{\Omega} |f-h|^p\,d\mu = \int_{\Omega-A} |f-h|^p\,d\mu +\int_{A} |f-h|^p\,d\mu $$ By Minkowski inequality $$ \int_{A} |f-h|^p\,d\mu <\left(\|f\|_{p}+\|h\|_{p}\right)^{p}<\delta^{1/p}(\|f\|_{\infty}+\|h\|_{\infty}) $$ for $f$ is $L_p$-integrable and $h^p$ is bounded. So let $\delta <(\frac{\epsilon}{2\|f\|_{\infty}+2\|h\|_{\infty}})^p$, there is $$ \int_{\Omega} |f-h|^p\,d\mu<\frac{\epsilon}{2\mu(\Omega)}\mu(\Omega)+\delta^{1/p}(\|f\|_{\infty}+\|h\|_{\infty})<\epsilon $$