If $f$ is a positive function and $\int_{E}f d\lambda = 0$ then $\lambda (E) = 0$

Question

If $f$ is a positive function and $$ \int_{E}f d\lambda = 0, $$ then show that $\lambda (E) = 0$ where $\lambda $ is a Haar (Radon) measure.

I know that if $f$ is a positive function and $\int_{E}f d\lambda = 0$ then $\lambda ^ \ast = 0 $ ($\lambda ^ \ast$ is the outer measure) but I can't show that $\lambda (E)$ is zero. Note: $E\subseteq G$ and $G$ is topological group.

Answer 1

Let $$ E_1=\{x\in E: f(x)\le 1\}, $$ and $$ E_n=\{x\in E : 2^{-n+2}> f(x)\ge 2^{-n+1}\}. $$ Clearly, the $E_n$ are disjoint, measurable, $E=\cup_{n=1}^\infty E_n$ and thus $$ \lambda(E)=\sum_{n=1}^\infty \lambda(E_n). $$ It suffices to show that $\lambda(E_n)=0$, for all $n$. But as $f(x)>0$, for all $x\in E$, $$ \int_E f\,d\lambda \ge \int_{E_n}f\,d\lambda\ge\int_{E_n} 2^{-n+1}d\lambda=2^{-n+1}\lambda(E_n), $$ which implies that $\lambda(E_n)=0$.

Answer 2

On that page of Aliprantis, $\mu$ is just a set-function, and $\mu^*$ is the outer measure it generates. Indeed $\mu^*(E) \le \mu(E)$ in that setting, and equality may not hold. For example, in the real line, $\mu(E)$ could be the diameter of the set $E$, and then for $\mu^*$ we get Lebesgue outer measure.

Of course when we write $\int_E f\;d\lambda$ we assume $\lambda$ is a measure, not just any set-function.