I haven't had measure theory, but now I stumbled on integration with respect to a specific measure, and now I'm confused.

$$\int \mathrm{d}x$$ or $$\int \mathrm{d}\mu(x)$$

What is meant here is that the infinitesimal amount $\mathrm{d}x$ implicitly understood as the Lebesgue measure of $x$ can be understood with any kind of measure $\mu$. How does this reflect the integration, if I, for instance, want to integrate $x^2$ with respect to another measure other than Lebesgue?

Of course, I've tried to google my way through, but the basic notion is confusing to me.

  • 9
    $\begingroup$ Let me describe a favourable situation to encourage. Not always but quite often a measure $\mu$ on $\langle\mathbb R,\mathcal B\rangle$ has a density with respect to the Lebesguemeasure. If that is the case and $f$ is such a density then $\int x^2d\mu(x)=\int x^2f(x)dx$. Btw, not that I want to increase your confusion now, but personally I would write $\int x^2\mu(dx)=\int x^2f(x)\lambda(dx)$ where $\lambda$ denotes the Lebesguemeasure. Somehow the infinitesimal amount $dx$ is "measured". On LHS by $\mu$ and on RHS by $\lambda$, and this with $\mu(dx)=f(x)\lambda(dx)$. $\endgroup$
    – drhab
    Jul 26, 2017 at 11:50
  • $\begingroup$ Could you explain density in this case? It works as a sort of mapping from whatever measure we are dealing with to the lebesgue measure as far as I understand on your notation. $\endgroup$
    – 1233023
    Jul 26, 2017 at 12:08
  • $\begingroup$ Look up Lebesgue integral. $\endgroup$
    – user223391
    Jul 26, 2017 at 12:16
  • $\begingroup$ @1233023 On base of the Lebesgue measure together with a suitable function $f$ you can define a new measure $\mu$ by stating that $\mu(A):=\int_Af(x)dx$ on measurable sets. This works through in integration and leads to $\int g(x)d\mu(x)=\int g(x)f(x)dx$ for suitable functions $g$. Here $f$ is the density of $\mu$ wrt the Lebesgue measure. If you know this density of $\mu$ and you know how to integrate wrt Lebesgue measure then also you know how to integrate wrt to $\mu$. $\endgroup$
    – drhab
    Jul 26, 2017 at 12:42
  • $\begingroup$ Thank you, very helpful! $\endgroup$
    – 1233023
    Jul 26, 2017 at 12:57

1 Answer 1


First, let's define what is a measure:

Given a class $\mathscr{F}$ of subsets of a set $\Omega$, a measure:


is a function having the following properties:

  1. $\mu(A)\geqslant 0$ all A in$\mathscr F$

  2. $\mu(\emptyset)=0$

  3. For a countable collection $A_j\in \mathscr{F},j\in\mathbb{N}$ with $A_j\bigcap A_j'=\emptyset$ for $j\neq j'$ and $\bigcup_\limits{j}^{}A_j\in F$

  4. $\mu \left(\bigcup_\limits{j}^{}A_j \right)=\sum_\limits{j}^{}\mu(A_j)$

Lesbegue measure is the length measure and it is you usually defined as $\lambda$. If we take an arbitrary set $[a,b]$, $\lambda[a,b]=b-a$. If you want you can see the Lebesgue measure actually fits the definition of a measure.

If you recall the Riemmann integral can be written as $\int_\limits{a}^{b}f(x)dx=\lim_{{\triangle x}\to 0}\sum_\limits{j}^{} f(x_j)\times\triangle x_j$ in which $\triangle x_j$ can be thought of the subsets of the interval $[a,b]$.

If you want to compute the integral using the Lebesgue measure you have:

$\int_\limits{a}^{b}fd\lambda=\sum_\limits{j}^{} f\mathbb{1}_{[a_j,b_j]}\lambda=\sum_\limits{j}^{} f(b_j-a_j)$ in which $\mathbb{1}$ is the indicator function taking the value one if the function takes value on the specific partition of $[a_j,b_j]$ and 0 otherwise.

Therefore now you do not require function to be continuous.

The difference between Lebesgue and Riemann integral is that you no longer need to make use of the infinitesimal $dx$ instead you break the function into small partitions. If the function is continuous the Lebesgue and Riemann integral coincide.

You can have the integral you like if you change the measure. Look at the Dirac functional, it is a clear example. I provide you the following example which was an answer given to me in another thread:

Dirac measure $\delta_{x}$ at $x$ is a Borel measure defined by $$ \delta_x(E)=\begin{cases} 1, & \mbox{ if }x\in E\\ 0, & \mbox{ if }x\notin E \end{cases}, $$ where $E\subseteq\mathbb{R}$ is a Borel set.

Given a Borel function $f:\mathbb{R}\rightarrow\mathbb{R}$, we have $$ \int f\,d\delta_{x}=\int_{\{x\}}f\,d\delta_{x}+\int_{\{x\}^{c}}f\,d\delta_{x}=f(x)\delta_{x}(\{x\})+0=f(x). $$


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