= Solution
The Gaussian function is continuous and hence <Lebesgue measurable>. For $|x|\geq2$, $x^2/2\geq|x|$, and therefore
$$
0\leq e^{-x^2/2}\leq\mathbf1_{[-2,2]}(x)+e^{-|x|}.
$$
The right side has finite <Lebesgue integral>: its first term has integral 4 and the second is <Lebesgue integrable> by hypothesis. The comparison principle for the <Lebesgue integral> says that a measurable nonnegative function dominated by an <integrable function> has finite integral. Consequently \b[the Gaussian function is <Lebesgue integrable>].
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