Can every measure be normalized in order to be a probability measure?

Question

Let $\mu$ be a measure on $(X,\mathcal{A})$. Is it possible to normalize $\mu$ in order to get a probability measure?

My idea is to set $$ \mu'(A):=\mu(A)/\mu(X)~\forall~A\in\mathcal{A}. $$

Answer 1

As pointed out in the comments, if $\mu$ is a finite measure on $X$ (i.e. $\mu(X) < \infty$), then you can normalise $\mu$ as you suggested to obtain a probability measure on $X$. If $\mu$ is not a finite measure on $X$ (i.e. $\mu(X) = \infty$), then you cannot do this. If $\mu$ is $\sigma$-finite (i.e. there are disjoint $X_n$, $n \in \mathbb{N}$, with $\mu(X_n) < \infty$), then one can define a probability measure by

$$\mu'(A) = \sum_{n=1}^{\infty}\frac{\mu(A\cap X_n)}{2^n\mu(X_n)}$$

but this concentrates the measure in $X_1$ more than $X_2$ and so on (e.g. $\mu'(X_n) = \frac{1}{2^n}$, regardless of the relative size of $\mu(X_n)$).

An example where you would like to create a probability measure but can't is the natural numbers with the counting measure. One reason for this is that you would like to answer questions such as: what is the chance a random integer from $\mathbb{N}$ is even? As is explained in the answers to that question, there is no countably additive probability measure on $\mathbb{N}$ which assigns equal measure to singletons (any such measure would simultaneously be a multiple of the counting measure and satisfy $\mu(\mathbb{N}) = 1$). In this situation, we instead use the notion of natural density; unfortunately, not every subset of $\mathbb{N}$ has a natural density.