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Let $(E, A)$ be a measurable space and $\delta_x$ be the dirac measure for $x \in E$ Prove that $\int ud \delta_x = u(x) \ \forall u \in M_{\mathbb{R \cup \{-\infty, \infty\}}}(A)$

Here is my attempt. $u = u^{+} - u^{-}, \ u^{+}, u^{-} \in M_{\mathbb{R \cup \{-\infty, \infty\}}}(A)$. Thus $\int u d\delta_x = \int u^{-} + \int - u^{-}d\delta_x$. Now lets study $u^{+}$

$$ \int u^{+} d \delta_x = \sup\{I_{\delta_x}(g) : g \in \epsilon^{+}(A), \ 0 \leq g \leq u^{+}\} \\ I_{\delta_x}(g) = \sum_m y_m \delta_x (A_m) \\ g = \sum_m y_m 1_{A_m} \in \epsilon^{+}(A). $$ Now since all $A_m$ are disjoint $\exists !m$ s.t $x \in A_m$, call this $m$ for $m_0$. We knos this $A_{m_0}$ exists since $x \in E = \cup_m A_m$. Thus

$$ I_{\delta_x} (g) = \sum_m y_m \delta_x (A_m) = y_{m_0} \delta_x (A_{m_0})= y_{m_0}, $$ thus $\int u^x d\delta_x = u^+(x) \Rightarrow \int u d \delta_x = u(x)$. Because we can do the same for $u^-$ and $u = u^+ - u^-$.

Is this correct?

1 Answers 1

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Observe that $$\int_E u\text{d}\delta_x=\int_{\{x\}}u\text{d}\delta_x=u(x)\int_{\{x\}}\text{d}\delta_x=u(x)\delta_x(\{x\})=u(x),$$ since $u$ is constant on a singleton $\{x\}$. Here I use the form of a Dirac measure: $$\delta_x(B)=\chi_B(x)=\begin{cases}1&\text{if }x\in B,\\ 0&\text{if }x\not\in B.\end{cases}$$

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    Sure! That was a nice proof. But is my proof correct? – 2017-02-18
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    I did not read it in detail because I am so lazy and your proof is too long for me to read :-). So I preferred to write my own :-) I will check your argument in a moment. In fact your argument is the same that I gave. But its construction is too complicated. This is the same in a place when you use $m_0$ and so on...\ – 2017-02-18
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    Do you mean that $u$ is nonzero on a singleton $\{x\}?$ – 2017-02-18
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    Does not matter, what is $u(x)$, zero or not zero. If zero, the integral is zero. – 2017-02-18
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    Yes, but it's what I came up with haha, we all think different. – 2017-02-18