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In the context of measure theory, given a probability measure $\xi : \mathcal{B}(X) \rightarrow [0,1]$ and a (smooth) function $v:X\rightarrow \mathbb{R}$ where $X\subset \mathbb{R}^n$, we encounter the notation

$$ \int_X v(x)\, \xi(dx) $$

What is behind the notation $\xi(dx)$ formally ? Do probability measures act on differential forms ? If so, what is this action formally ?

Smilia
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  • If you have a differential $df$ then $\mu(df) = |\det df_x| \mu(dx) = |\det df_x| d\mu(x)$. In the case of $f:\mathbb{R}^n\to\mathbb{R}^m$ you have $|\det df_x| = \sqrt{\det G_f(x)} = \sqrt{\det\bigl((J_f(x)^T J_f(x)\bigr)}$. It basically measures the change in volume that $f$ induces infinitesimally (how much the best linear approximation $df$ of $f$ stretches or shrinks space). In fact $\sqrt{\det G_f(x)}$ is the volume of the parallelotope with edges the partial derivatives, i.e. the columns of $J_f(x)$. You also have $\mu (df) = d(\mu \circ f) = df^*\mu$. – lightxbulb May 21 '23 at 06:33

1 Answers1

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If $v$ is a measurable function, the integral in your question is just a notation for the Lebesgue integral of the function $v$ with respect to the measure $\xi$. Other standard notations are $$ \int_{X} v(x) \, \xi(dx):= \int_{X} v(x) \, d\xi(x) := \int_{X} v \, d\xi. $$

user980341
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    how is it related to integral of differential forms coming from differential geometry ? – Smilia Jan 06 '17 at 16:51
  • As already pointed out, $\xi(\text{d}x)$ is a notation to emphasize with respect to which measure the integral is taken and basically this has nothing to do with differential forms. – user980341 Jan 06 '17 at 22:11
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    For the relation of differential forms with measures see for example https://en.wikipedia.org/wiki/Differential_form#Relation_with_measures or these questions: http://math.stackexchange.com/questions/57606/integral-of-differential-form-and-integral-of-measure http://math.stackexchange.com/questions/49641/integration-of-forms-and-integration-on-a-measure-space?noredirect=1&lq=1 – user980341 Jan 06 '17 at 22:14