# Can we always find a measure for a measurable space?

I'm new to the Measure Theory. I was wondering can we always find a measure for a measurable space? It would be better to explain in details.

• There's always the measure that maps any non-empty set to infinity. Commented Sep 23, 2017 at 17:34
• The usual definition of a "measure space" is that it's a set of points with a measure. Do you have something else in mind when you ask about a "measurable space"? Commented Sep 23, 2017 at 17:39
• The hardest measurable space over $X$ to assign a measure function is always the power set of $X$. Why? Maybe this helps your intuition. Commented Sep 23, 2017 at 17:40
• This is an old question and I am adding my comment because I have the exact same doubt. The origin of my doubt is from the statement "There is no translationally invariant uniform probability measure on power set of reals." That made me think "are there any possible measure possible on an arbitrary sigma algebra?" Motivation to define a sigma algebra is to make sure that domain of a measure will make sense when we define measures. Commented Feb 1, 2019 at 22:18
• However, if we have a meaningful domain in the form of a sigma algebra, does that necessarily imply that it is always possible to define a non-trivial measure (need not be a probability measure but some measure)? Commented Feb 1, 2019 at 22:18

## 3 Answers

There's always the measure that maps every subset to $0$.

If you mean a probability measure then it also exists, pick $x\in X$ and define $f(A)=1$ if $x\in A$ and $0$ otherwise. This is called a Dirac measure.

• One may add that the counting measure is also defined on any measurable space. Commented Sep 23, 2017 at 17:44

A measurable space is just a pair $(X,\mathcal{M})$, where $X$ is a set and $\mathcal{M}\subseteq\mathscr{P}(X)$ is a $\sigma$-algebra on $X$. The purpose of this definition is just to identify a $\sigma$-algebra on $X$, which is the collection of all measurable subsets of $X$.

For any given set $X$, there are several ways in which we can define a measure $\mu:\mathscr{P}(X)\to[0,\infty]$. (See examples below.) For any such measure, the restriction $\mu|_{\mathcal{M}}$ will turn the measurable space $(X,\mathcal{M})$ into the measure space $(X,\mathcal{M},\mu|_{\mathcal{M}})$. In fact, it's worth proving the following easy exercise.

If $(X,\mathcal{M},\mu)$ is a measure space and $\mathcal{N}\subseteq \mathcal{M}$ is a $\sigma$-algebra, then $\mu|_{\mathcal{N}}$ is a measure on the measurable space $(X,\mathcal{N})$.

Some important examples of measures $\mu:\mathscr{P}(X)\to[0,\infty]$:

1. The zero measure: $\mu(A)=0$ for every $A\subseteq X$.
2. The infinite measure: $\mu(A)=\infty$ for $A\ne\emptyset$ and $\mu(\emptyset)=0$.
3. The counting measure: $$\mu(A) = \begin{cases} |A| & \text{if A is finite}, \\ \infty & \text{otherwise}. \end{cases}$$
4. Point mass measure (or Dirac measure) of a fixed point $x\in X$: $$\mu(A) = \begin{cases} 1 & \text{if x\in A}, \\ 0 & \text{otherwise}. \end{cases}$$
5. $\mu(A)=\infty$ if $A$ is infinite, and $\mu(A)=0$ otherwise.

To construct a probability measure on any measurable space $$(\Omega,\tau)$$, consider the Dirac measure P: For any $$A\in \tau$$ $$P(A) = \begin{cases} 1 & \text{A\ni x}, \\ 0 & \text{otherwise}. \end{cases}$$ Choice $$x\in \Omega$$ such that: $$\exists B\in \tau, B\ni x$$. Then $$(\Omega,\tau,P)$$ is a probability space, P is a probability measure.