Showing a extended real valued function is measurable, problem dividing by $0$ and $\infty$

Question

Suppose $f(x)= \frac{1}{x}$ if x is not $0$, and $f(x)= \infty$ if $x=0$. How do I show that f is borel measurable? I was thinking of writing f as $f(x)= \frac{1}{x} 1_{\{x \neq 0\}} (x) + g(x) 1_{\{x=0\}}$ where $g(x)= \infty$ for all x. Then as $\{(a,\infty]: a \in \mathbb{R}\}$ is a set of generators for the borel sigma algebra of the extended reals, clearly 1/x is borel measurable, and so is each indicator and finally g as well as the inverse image of any such interval is all real numbers.However, what bothers me is the fact that $\frac{1}{x} 1_{\{x \neq 0\}} (x)$ does not even make sense at $x=0$ as the function $\frac{1}{x}$ is undefined there. Also what bothers me is g(x) will be infinity when x is not 0 but the indicator will be 0, and we will have $\infty$ times 0. Could anyone clear things up for me?

Answer 1

You show something more general, namely that if you have 2 functions $f$ and $g$ ,$f$ being mesurable and $μ([g\neq f])=0$ (the points that $g$ differs from $f$ has measure 0) then g is also measurable. To do this compare the sets $[f\leq a]$ and $[g\leq a]$. Use the fact that Lebesgue measure is complete, namely that if a you have two sets $A$ measurble and $B$ and $μ(A\bigtriangleup B)=0$ then $B$ is measurable.

Answer 2

However, what bothers me is the fact that $\frac{1}{x} 1_{\{x \neq 0\}} (x)$ does not even make sense at $x=0$ as the function $\frac{1}{x}$ is undefined there. Also what bothers me is $g(x)$ will be infinity when $x$ is not $0$ but the indicator will be $0$, and we will have $\infty$ times $0$.

Note that you are actually using $f(x) 1_{\{x \neq 0\}} (x)$ here, so it will be well-defined: in the space $\overline{\mathbb{R}}=[-\infty,\infty]$ we have by definition (this definition is usual in measury theory) that $0\cdot \infty=0$, so both $f(x) 1_{\{x \neq 0\}} (x)$ and $g(x) 1_{\{x = 0\}} (x)$ are well-defined.