Let $f$ be a Lebesgue measurable function defined on a set $E\subset \mathbb{R}^d$. Let $\{x\in E\mid f(x)\neq 0\} $ be the support of the function on $E$.
Stein and Shakarchi claim without proof that the support is measurable. But why is this true? It's not just due to the fact that the support is a subset of a measurable set.
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A function $f:\mathbb{R}^d\mapsto \mathbb{R}$ is Lebesgue measurable by definition if $$\{x\in \mathbb{R}^d \ | \ f(x) > a \}$$ is measureable for all $a\in \mathbb{R}$. Equivilently you may use $\leq, <$ or $\geq$ instead of $>$. Therefor $$\{ x\in \mathbb{R}^d \ | f(x) \neq 0 \} = \{x\in \mathbb{R}^d \ | \ f(x) > 0 \}\cup\{x\in \mathbb{R}^d \ | \ f(x) < 0 \} $$
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As a general rule, anything that you can define explicitly in terms of a measurable function or sequence of functions is going to turn out to be measurable; it takes a little trick to show that not every set is measurable.
That's too broad to be provable, but this particular example is trivial. Say $f=u+iv$ where $u$ and $v$ are real-valued. Then $\{f(x)\ne0\}=\{u(x)\ne0\}\cup\{v(x)\ne0\}$. So you can assume wlog that $f$ is real-valued. So $\{f(x)\ne0\}=\{f(x)>0\}\cup\{f(x)<0\}$, the union of two measurable sets.
The support is usually defined as the closure of that set, but I'll answer your specific question.
Note that your set is precisely $$f^{-1} ((-\infty,0)\cup (0,\infty))=f^{-1}((-\infty,0))\cup f^{-1}((0,\infty)).$$ Now, just recall the definition of a Lebesgue measurable function.