Notation measure theory: $d\mu(x)$

Question

Let $\mu$ be a measure on a measurable space $(S,\mathcal{A})$ if we have the integral $\int_S f(x)d\mu(x)$ with $f$ a real valued function, what does $d\mu(x)$ mean? Why is the differential a function of x, what is the intuition here?

Answer 1

Let's think back to the regular calculus integral: $\int f(x)\,dx$. What does the $dx$ mean here? Officially, $\int f(x) \,dx$ is the number you get when you take the infimum of the upper sums over all partitions (or the supremum of the lower sums over all partitions -- the sup and inf should be the same for the Riemann integral to exist).

The above definition is a technical definition; however, we can intuitively think of $\int_{a}^{b} f(x)\,dx$ as an infinite sum $\sum\limits_{\alpha_{i}} f(\alpha_{i})\cdot \Delta_{i}$, where $\alpha_{i}$'s represent all real numbers between $a$ and $b$, and $\Delta_{i}$ is the width of an infinitely thin rectangle whose base contains the $x$-value $\alpha_{i}$. (So the intuition here is to partition the domain/$x$-values, and add up over all $x$-values. This will be different in the Lebesgue integral below.) Note that in this intuition, the infinite sum would be an uncountable sum, not the usual countable sum you learned about in Calculus. We don't really have a way to sum up uncountably many items, so instead we take the supremum (or infimum). So in the Riemann integral case, $dx$ is the length of the base of an infinitely thin rectangle containing $\alpha_{i}$ for each $x=\alpha_{i}$ in $[a,b]$.

In the two paragraphs above, I tried to highlight the actual technical definition, and then the intuition which motivates the definition. In the same way, we can discuss $\int f(x)\,d\mu(x)$. With the Lebesgue integral, if $f\geq 0$, the official definition of $\int f(x)\,d\mu(x)$ is the number you get when you take supremum of the the Lebesgue integrals of all nonnegative simple functions $s(x)$ sandwiched between $0$ and $f$, i.e., where $0 \leq s \leq f$. The Lebesgue integral of the simple function $s(x) = \sum_{i} c_{i}\Bbb 1_{A_{i}}$ is $\int s(x)\,d\mu(x) = \sum_{i} c_{i}\Bbb \mu(A_{i})$. So taking the supremum of these integrals over all simple functions sandwiched between $0$ and $f$ gives you a number which we define $\int f(x)\,d\mu(x)$ to be (for $f \geq 0$).

Intuitively, $\int f(x)\,d\mu(x)$ is also an uncountable sum. However, instead of adding up terms over all possible values of $x$ as in the Riemann integral case, we add up over all possible values of the variable $y$ (where $y = f(x)$). So let's say $y = f(x)$ will take output values in the interval $[c,d]$. We want to add up terms that look like: $y_{i} \cdot \mu(\{x \mid f(x) = y_{i}\})$, i.e., possible value $y_{i}$ of $f$, multiplied by the size of the domain subset where $f$ will equal $y_{i}$. So $\int f(x)\,d\mu(x) = \sum_{y_{i}} y_{i}\mu(\{x \mid f(x) = y_{i}\})$. Of course, this is only intuition, since, again, we don't have a notion of uncountable sum in math. So $d\mu(x)$, which is also often noted $\mu(dx)$, represents the measure of the set of $x$ such that $f(x)$ equals $y_{i}$.

Hope this helps!

Answer 2

$\int_S f(x)d\mu(x)$ is just another notation for $\int_S f d\mu$. This notation is sometimes useful when there are several variable involved. For example $\int_S f(x,y) d\mu (x)$ denotes the integral of the function $g(x) =f(x,y)$ w.r.t. $\mu$ when $y$ is fixed.