Let $(\mathcal{X},\mathcal{M},\nu)$ a measure space where $\nu$ a signed measure. Show that there exists a measurable function $g$ with $|g| \leq 1$ so that $\int_E g d\nu = |\nu|(E)$ for all measurable sets $E$.

Here's my attempt.

Let $E$ a measurable set. Then $|\nu|(E) = sup\sum_{j=1}^\infty |\nu(E_j)|$. Let $\epsilon >0$ Then i can always find a partition $E=\bigcup_{i=1}^\infty G_i$ s.t. $\sum_{i=1}^\infty |\nu(G_i)| \geq |\nu|(E)-\epsilon$. Let $\chi'_{G_i} = sign(\nu(G_i))\chi_{G_i}$ where $\chi_{G_i}$ the characteristic function of set $G_i$. Let $g=\sum_{i=1}^\infty \chi'_{G_i}$. Then $\int_E gd\nu=\sum_{i=1}^\infty |\nu(G_i)|$. Thus $|\nu|(E)\geq \int_E gd\nu \geq |\nu|(E) -\epsilon$, and since e is chosen arbitrary $\int_E gd\nu = |\nu|(E)$.

Is my approach correct? Thank you very much for your help!

  • $\begingroup$ No, it doesn't work this way. Your function $g$ depends on $\epsilon$, i.e. you show that there exists $g_{\epsilon}$ such that $$|\nu|(E) \geq \int g_{\epsilon} \, d\nu \geq |\nu(E)|-\epsilon.$$ Therefore, you cannot conclude $\int_E g \, d\nu = |\nu|(E)$. $\endgroup$ – saz Apr 30 '15 at 10:45
  • $\begingroup$ Could anyone suggest a way to solve this ? $\endgroup$ – SpawnKilleR Apr 30 '15 at 15:59

As I already pointed out in a comment, your proof is not correct because the function $g$ does depend on $\epsilon$ and $E$. The following counterexample shows that the claim does not even hold true:

Consider $(\{-1,1\},\mathcal{P}(\{-1,1\}))$ endowed with the measure $$\nu(dy) =\delta_1(dy) - \delta_{-1}(dy).$$

Suppose that there exists a function $g$ such that $\int_E g \, d\nu = |\nu|(E)$ for all measurable sets $E$. For $E=\{\pm 1\}$, we have

$$g(\pm 1) \stackrel{!}{=} |\nu|(\{\pm 1\})=1,$$

i.e. $g(1) = g(-1) = 1$. On the other hand, for $E := \{-1,1\}$,

$$\int_E g \, d\nu = 1 \cdot g(1) + (-1) \cdot g(-1) = 0$$

does not equal


Remark: There exists a function $g$ such that $$\nu(E) = \int_E g \, d|\nu|.$$ This follows directly from the fact that $\nu$ is absolutely continuous with respect to $|\nu|$ and the Radon-Nikodým theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.