Is $\mathcal{M}(\mathbb{R}^d)$ composed of finite measures?

Question

Let $C_0(\mathbb{R}^d)$ be the space of real-valued continuous functions on $\mathbb{R}^d$ that vanish at infinity. By Riesz–Markov–Kakutani representation theorem, the dual of $C_0(\mathbb{R}^d)$ is $\mathcal{M}(\mathbb{R}^d)$ composed of all the real-valued regular Borel measures on $\mathbb{R}^d$. I am wondering whether all the measures in $\mathcal{M}(\mathbb{R}^d)$ are finite. I think it is true since by the definition of TV norm, for any $\mu\in\mathcal{M}(\mathbb{R}^d)$, $$ |\mu(\mathbb{R}^d)|\leq|\mu|(\mathbb{R}^d)=\|\mu\|_{\mathrm{TV}}<\infty. $$ But I check the textbook, for example "a course in functional analysis" by John Conway and it seems like the "finiteness" is not mentioned. So is the above argument correct?

Answer 1

Most authors take it as part of the definition of "real-valued measure" or "signed measure" that the measure should be finite, i.e. that $\mu(A)$ is a finite real number for each set $A$ in the $\sigma$-algebra. Certainly that is what's intended here.

Some condition like this is needed in order for the additivity property of the measure to make sense. If $A,B$ are two disjoint sets, we are supposed to have $\mu(A \cup B) = \mu(A) + \mu(B)$, but if we had $\mu(A)=+\infty$ and $\mu(B)=-\infty$ then it doesn't make sense to even write $\mu(A) + \mu(B)$. By requiring the measure to be finite-valued the issue doesn't arise.

You can also avoid this issue by considering measures that may either take the value $+\infty$ or $-\infty$ but not both. But it is still not too satisfactory, as the set of such measures does not form a vector space.