Past exam of the options I course of the University of Oxford / 2006 / D / D1 / i

The Gaussian function is continuous and hence Lebesgue measurable. For $|x|\ge 2$, $x^2/2\ge |x|$, and thereforeThe 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 the Gaussian function is Lebesgue integrable.