In my professor's notes on Lebesgue Integration, the proof of the Theorem above says that we need to prove that for all measurable sets $E$ with finite measure, and for $\epsilon>0$ exists a function $g_\epsilon$, continuous with compact support, such that $\Vert\chi_E - g_\epsilon\Vert < \epsilon$. There exists a compact $K$ and open $G$ such that $K\subseteq E\subseteq G$ and $\lambda(G - K)<\frac{\epsilon}{2}$. If we find $g$ with $supp(g)\subseteq G$ and $\chi_K\leq g\leq 1$ then $$\Vert\chi_E - g\Vert = \int_K |\chi_E-g|d\lambda + \int_{G-K} |\chi_E-g|d\lambda + \int_{\mathbb{R}^n-G} |\chi_E-g|d\lambda$$ Because $\chi_E=1$ in $K$ and $1=\chi_K\leq g\leq 1$, then $g=1$, and the first integral is $0$, and because the support is contained in G, then $g$ is $0$ a.e. in $\mathbb{R}^n-G$ and evidently $\chi_E=0$ in this set, so the integral is 0 too. Then we get that $\Vert\chi_E-g\Vert\leq2\lambda(G-K)<\frac{\epsilon}{2}.$

My question: Where are the hypoteses of $g$ being continuous and with compact support used?

Thanks in advance.


1 Answer 1


The hypothesis isn't used anywhere.

Your proof only uses the hypothesis that $g$ is measurable, that $g|_K\equiv 1,$ $g|_{\mathbb{R}\setminus G}\equiv 0$ and $\|g\|_{\infty}\leq 1$.

Hence, the real question is: Does such a $g$ exist, if we also require $g$ to be continuous with compact support? I.e., given $K\subseteq G$ with $G$ open and $K$ compact, does there exist a continuous function with compact support such that $f|_K\equiv 1$ and $f|_{\mathbb{R}\setminus G}\equiv 0$? This is a special case of Urysohn's Lemma in locally compact spaces.

  • $\begingroup$ Oh, so i see that adding the restriction of a continous and with compact support $f$ still can be found, but these properties are anywhere used to complete the proof. I was having difficulties understanding it because then the proof proceeds to find such a function with those properties, but they aren't seem to be used. $\endgroup$
    – deiv
    Jan 22, 2020 at 15:02

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