Square Integrable Functions are Measurable?

Question

I have a simple and potentially stupid question but somehow I just can't get it. In Tom Apostol's Mathematical Analysis, Theorem 10.54 it says that if $f$, $g$ are square integrable then $f$ and $g$ are measurable. It looks plausible but may I know why? While $f^2$ and $g^2$ are Lebesgue-integrable and hence measurable, that I can understand, I don't see why it can be extended to $f$ and $g$ directly... Thanks! Bosco

Answer 1

As stated, it is false. If $V$ is non-measurable subset of $[0,1]$ and if $f\colon V\longrightarrow\mathbb R$ is the function defined by$$f(x)=\begin{cases}1&\text{ if }x\in V\\-1&\text{ otherwise,}\end{cases}$$then $f^2$ is Lebesgue-integrable, but $f$ is not measurable.