I suspect the following is true, but I don't know a reference. Let $X$ be a measurable space (with $\sigma$-algebra $\mathcal{F}(X)$), and let $MX$ denote the space of finite signed measures on $X$. Then \begin{equation*} \phi(\nu)=\int_X\phi(\delta_x)\,d\nu(x), \end{equation*} for all $\phi\in(MX)'$ and $\nu\in MX$, where $\delta_x\in MX$ is the Dirac measure concentrated at $x\in X$.
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See here and here for more information on what these functionals are. They are quite unruly objects. – Michael Greinecker Oct 16 '22 at 20:15
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No. Let $X=[0,1]$ with the Borel sets. If $\mu$ is atomless, let $\lambda(\mu)=0$. If $\mu=\sum_{i=1}^\infty\alpha_i\delta_{x_i}$ is discrete, let $\lambda(\mu)=\sum_{i=1}^\infty\alpha_i$. Extend $\lambda$ by linearity to all finite signed measures.
For $\mu$ the uniform distribution, we have $$\lambda(\mu)=0\neq 1=\int 1 \mathrm d\mu=\int \lambda(\delta_x)~\mathrm d\mu(x).$$
Michael Greinecker
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Hi @Michael, thanks for the example. I have one question about it. Surely the set of discrete measures (with finite support) forms a dense subspace of the space of all measures on $[0,1]$. In that case, the functional given by \begin{equation}\lambda\biggl(\sum_{I=1}^n\alpha_i\delta_{x_i}\biggr)=\sum_{I=1}^n\alpha_i\end{equation} on the space of discrete measures has a unique extension to a continuous functional on the space of all measures on $[0,1]$. – HardyHulley Oct 17 '22 at 16:31
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The discrete measures are weak*-dense, where this applies. They are not norm dense in the variation norm. You are fine if you ude the Gelfand integral insted of the Pettis integral. – Michael Greinecker Oct 17 '22 at 16:41
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