I know and I can prove that, given $f:\Omega \rightarrow \mathbb{R}$ in $L^1_{|\mu|}(\Omega)$, then $$\left|\int_{\Omega}f\,d\mu\right|\leq \int_{\Omega}|f|d|\mu|$$ where $|\mu|$ is the total variation measure of $\mu$. But the same holds even if $f:\Omega \rightarrow \mathbb{R}$ and $\mu$ is a vector-valued Radon measure in $\mathbb{R}^d$ and $\int_{\Omega} f\,d\mu=(\int_\Omega fd\mu_1,...,\int_\Omega f d\mu_d)$. How can I prove it?
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?? What are $\mu_1, \mu_2, \dots$? Shouldn't it be $(\int_\Omega f_1,d\mu, \int_\Omega f_2,d\mu, \dots)$? – Paul Sinclair Dec 26 '23 at 16:10
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$\mu$ is a vector Radon measure in $\mathbb{R}^d$ while $f$ is not vector-valued. – nimaba99 Dec 27 '23 at 18:51
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You said "$f:\Omega \to \Bbb R^d$". How is that "not vector-valued"? – Paul Sinclair Dec 28 '23 at 03:09
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You're right, that was a typo. I'll fix it – nimaba99 Dec 29 '23 at 14:13
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related https://math.stackexchange.com/q/3696228/121671 – Mittens Dec 29 '23 at 16:28