measure defined in terms of an integral

Question

Here's a problem from a midterm I took 2 decades ago! Let $A$ be a Borel measurable set on the real line and $\lambda$ be the Lebesgue measure. Define a measure $\mu(A)=\int_A f d\lambda$ where $f(x)=1/x^2$ for $x\ne 0$ and $f(0)=\infty$.

1. Is $\mu$ finite? $\sigma$-finite?

Now I know it's not finite, since if $A$ is any interval containing $0$ then the integral is infinite. I'm not sure about $\sigma$-finite though. Any interval that does not contain $0$ will have finite measure, and there's a countable number of these to cover everything but $\{0\}$, so it seems to me that the answer depends on the following....

2. What is $\mu(\{0\})$?

Answer 1

$\mu$ is $\sigma$-finite. $$ \mathbb R = \{0\} \cup \bigcup_{n=2}^\infty [1/n,n] \cup \bigcup_{n=2}^\infty [-n,-1/n] $$ and all those sets have finite measure.

Answer 2

OK - I was thrown off by the facts that

1. $f$ takes values on the extended reals and
2. $f\chi_A$ is not necessarily integrable

but I should have gone back to basics. Here's my attempt at filling in the details:

Despite 1. and 2. above, $f\chi_A$ is still measurable for all Borel measurable sets $A$. In particular, $f\chi_{\{0\}}$ is Borel measurable, and can be approximated by simple functions $s_n$ such that $0 \le s_n \le f\chi_{\{0\}}$ (take $s_n = n\chi_{\{0\}}$ for example). Since the support of $s_n$ is a Borel set of measure $0$, $\int_{\mathbb{R}} s_n=0$ for all $n$, and $\int_{\mathbb{R}} f\chi_{\{0\}} = sup \int_{\mathbb{R}} s_n=0$. This shows that $\mu$ is absolutely continuous with respect to $\lambda$.