# Let $f\in \mathscr{L}^1$. Show that for every $\epsilon$ there exists a continuous function $g$ such that $\int_X |f-g|d\mu < \epsilon$.

Let $$f\in \mathscr{L}^1$$. Show that for every $$\epsilon$$ there exists a continuous function $$g$$ such that $$\int_X |f-g|d\mu < \epsilon$$.

Since the question is asking for a sequence of continuous functions that approximate $$f$$ in the integral, I'm thinking about applying Lusin's theorem and the monotone convergence theorem. Construct a sequence of increasing continuous functions $$\{f_n\}$$ such that $$f_n \to f$$. Then $$\int_X f_n d\mu \to \int_X f d\mu$$. Am I on the right track?

For example, let $$f_n=f-\frac{1}{n}$$. We have $$\int_X f_n d\mu \to \int_X f d\mu$$. By Lusin's theorem, there exists a continuous function $$g_n$$ such that $$\mu(\{x|f_n(x)\neq g_n(x)\})<\epsilon$$. But how can I relate $$g_n$$ to $$f$$ in the last step?

• This is a basic theorem in the theory of measure and integration. Try to read the proof from your favorite book and if you get stuck we would be glad to help you. May 21 '19 at 6:20
• Perhaps a relevant post here May 21 '19 at 6:45
• @KaviRamaMurthy Could you please recommend a text? I have read Rudin and Royden but they don't have this theorem May 21 '19 at 8:13
• They both have this result. In Rudin See the section "Approximation by continuous functions" in the chapter on $L^{p}$ spaces. May 21 '19 at 8:16