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This is maybe an obvious question but I'm struggling to find a source that clarifies this notation to me. In a paper I'm reading I find they define a measure $\mu_{\alpha}$ saying $d\mu_{\alpha}(x)=|x|^{\alpha} dx$. Later on, given another measure $\mu$ they define a sequence of measures $\nu_n$ by $d\mu_n=f_nd\mu$, where $f_n$ is a sequence of functions.

I guess this means that, when you use this measures to integrate, you replace the $d\mu_{\alpha}$ or $d\nu_n$ by their respective formulas. My question is: how can I get an explicit formula, if possible, for these measures? And, furthermore, why is it guaranteed that they're measures in the first place?

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If we say $d\mu$ is a measure, we really mean that $\mu$ is a measure defined by $$ \mu(E) = \int_E 1\,d\mu. $$ In other words, we obtain the measure of a set by integrating $1$ with respect to the measure $\mu$. If, for example, $d\nu = f\,d\mu$, then $$ \nu(E) = \int_E1\,d\nu = \int_E f\,d\mu. $$ This is just notation as it is.

Here is a simple exercise you should be able to do that establishes rigorously that such formulas define measures in certain cases. Suppose that on a measure space $(X,\mu)$, we are given a nonnegative function $f$ that is $\mu$-integrable, i.e., $\int_X f\,d\mu < \infty$. Prove that the set function $$ \nu(E) = \int_E f\,d\mu $$ defines a measure on $X$. You have to check the axioms of a measure are satisfied by $\nu$, and this will come down to applying the familiar limit theorems of Lebesgue integration.

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When you write $d \mu = f d \nu$ it means that $\mu$ is the measure defined by $$ \mu (A) = \int_A fd\nu $$ which gives you the formula explicitly.

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