If I have a measure $\mu$ on $[0,1]$ and if I know that

$\int_{[0,1]}Gd\mu\leq\int_0^1|G(r)|dr\quad \forall G\in C[0,1]$

this implies that the measure $\mu$ is absolutely continuous with respect the Lebesgue measure but I don't understand the reason.

I should prove that if $\lambda(A)=0$ then $\mu(A)=0$ where $\lambda$ is the Lebesgue measure. If I could replace $G$ with $\mathbb{I}_A$ it would be ok, but $G$ must be a continuous function and $\mathbb{I}_A$ is not a continuous function.

Thank you and sorry if my English wasn't correct.

  • $\begingroup$ Hint: Smooth functions with compact support are dense in $L^1(0,1)$. $\endgroup$ – user251257 Jul 22 '15 at 9:57
  • $\begingroup$ @user251257 Approximating $\mathbb{I}_A$ in the $L^1$ sense doesn't get you much, since $\mathbb{I}_A = 0$ almost everywhere ($A$ has Lebesgue measure 0). $\endgroup$ – Mike Haskel Jul 22 '15 at 9:59
  • $\begingroup$ @MikeHaskel, but that's the point. We want to show that $\mu(A) = 0$. $\endgroup$ – user251257 Jul 22 '15 at 10:02
  • $\begingroup$ @MikeHaskel: There is an assumption on $\mu$ in the question. $\endgroup$ – user251257 Jul 22 '15 at 10:04
  • $\begingroup$ @user251257 Yeah, retracted my silly question. $\endgroup$ – Mike Haskel Jul 22 '15 at 10:05

Notice that $\mu([0,1]) \le 1$.

Let $I=[a,b]\subseteq[0,1]$ be an interval. Let $\epsilon > 0$. Then consider the piecewise linear function $f_\epsilon$ with $f_\epsilon(I) = 1$ and $f_\epsilon([0,1] \setminus [a-\epsilon, b+\epsilon]) = 0$. Then, we have $f_\epsilon \ge \mathbb 1_I$ and $f_\epsilon \to \mathbb 1_I$ almost everywhere and in $L^1[0,1]$. Thus, it follows $$ \mu(I) = \lim_{\epsilon\to 0+} \int_0^1 f_\epsilon \;\mathrm d\mu \le \lim_{\epsilon\to 0+}\int_0^1 f_\epsilon \;\mathrm d\lambda = \lambda(I). $$ That is, $\mu(I)\le \lambda(I)$ for any interval in $[0,1]$.

Let $A\subseteq[0,1]$ with $\lambda(A)=0$ and $\epsilon > 0$. Now, cover $A$ by an union of intervals $\bigcup_{n=1}^\infty I_n$ with $\sum_{n=1}^\infty \lambda(I_n) < \epsilon$ (property of Lebesgue null set) and we have $$ \mu(A) \le \sum_{n=1}^\infty \mu(I_n) \le \sum_{n=1}^\infty \lambda(I_n) < \epsilon. $$

  • $\begingroup$ I think the first inequality is wrong. That's what was bothering me before. This answer is almost there, though. If $\mu$ is a positive (i.e., unsigned) measure, then we can take $f_\epsilon \geq \mathbb{I}_A$. If $\mu$ is signed, break it into positive and negative pieces. $\endgroup$ – Mike Haskel Jul 22 '15 at 10:11
  • $\begingroup$ @MikeHaskel you were right. I want to skip the technical proof with intervals and thought using convolution approximation would be easier. I hadn't thought it through. $\endgroup$ – user251257 Jul 22 '15 at 10:50


Prove that $\nu:=\lambda-\mu$ is a measure, wich comes to the same as proving that it takes nonnegative values.

Based on the data it can be shown that $\nu(I)\geq0$ for each interval $I\subseteq[0,1]$.

Then $\lambda A=0$ implies $\mu A=0$


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