# 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 $E$.

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!

• 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 $\int_E g \, d\nu = |\nu|(E)$. – saz Apr 30 '15 at 10:45
• Could anyone suggest a way to solve this ? – 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

$$|\nu|(E)=2.$$

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.