I'm reading measure theory and cannot find any definition of $d\mu$. The meaning of the measure $\mu$ over sets is always defined in all sources, but not that of $d\mu$. Why is the meaning of $d\mu$? Why is $d\mu=1$ for counting measure?
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The symbol $\mathrm d\mu$ only appears within integrals, as in $\int_Af\,\mathrm d\mu$. What it means is that we are integrating with respect to the measure $\mu$. And, by definition, saying that $\mu$ is the counting measure means $\mu(A)=\#A$; in particular, $\mu\bigl(\{x\}\bigr)=1$, for each $x$.
José Carlos Santos
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Typo: within, i – GNUSupporter 8964民主女神 地下教會 Oct 08 '18 at 07:24
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Thanks, but why $\int_A f d\mu$, and not $\int_{A,\mu}f$? Is it just convention? Or does the "d" mean something? – LyxNewbie Oct 08 '18 at 07:26
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@GNUSupporter8964民主女神地下教會 I've edited my answer. Thank you. – José Carlos Santos Oct 08 '18 at 07:31
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1@LyxNewbie, nothing special. Is simply a notation similar to $\int_a^b f(x),dx$. – Martín-Blas Pérez Pinilla Oct 08 '18 at 07:32
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@LyxNewbie It's a convention, but it is convenient to have something telling us what is the measure that we are working with. – José Carlos Santos Oct 08 '18 at 07:33
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Thanks, but it is still unclear. What if we have $\int_A f(x,y) d\mu(x) d\nu(y)$, where both $\mu$ and $\nu$ are counting measures? The "d" then doesnt seem just like a convention. – LyxNewbie Oct 08 '18 at 07:37
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2@LyxNewbie That $d$ is an historical relic from the days of the widespread use of infnitesimals. – José Carlos Santos Oct 08 '18 at 07:43
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You are correct, @LyxNewbie; it's a convention. The $d$ shouldn't be taken literally. Some mathematicians write $d\mu(x)$, others write $\mu(dx)$. The most correct would be to do something like you did: $\int_{A,\mu} f.$ – md2perpe Oct 08 '18 at 07:45
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Thanks, i get it now, it seems its a meaningless convention. Can someone add that its just a convention to some answer so i can accept it? – LyxNewbie Oct 08 '18 at 07:50
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@LyxNewbie Nobody except the post owner can add ideas to a post. – GNUSupporter 8964民主女神 地下教會 Oct 08 '18 at 07:59
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Personally I prefer the notation $\mu(f)$ where a measure is looked at as a function on functions instead of sets. See this question on that. – drhab Oct 08 '18 at 08:00
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The notations $\int_a^bf(x)\,\mathrm d\mu$ or $\int_a^bf(x)\,\mathrm d\mu(x)$ are (IMHO) not very helpful. A perhaps more informative equivalent notation is$$\int_a^bf(x)\mu(\mathrm dx).$$Let us think (rather coarsely) of $\mathrm dx$ as representing a small division of the interval $[a\;\pmb,\; b]$ about the point x, say $[x-\frac12\varepsilon\;\pmb,\; x+\frac12\varepsilon)$ of length $\varepsilon$. Then, in the simple case of the Riemann integral, the (inexplicit) measure applied to this interval is just its length $\varepsilon$, so that $\mu(\mathrm dx)=\varepsilon$. In the Riemannian case, which preceded the idea of measure, the "$\mathrm dx$" represents the (infinitesimal) length $\varepsilon$ of the interval, rather than the interval itself.
John Bentin
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Thanks, that was very helpful, if $d\mu = \mu(dx)$, it means you are somehow multiplying measure of a small set with the function, and the entire expression becomes a weighted average. So it seems its not merely a convention. – LyxNewbie Oct 08 '18 at 23:51