# Defining a measure $\mu$ by the function $d\mu$

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?

## 2 Answers

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.

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.