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

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.

A real measurable function is Lebesgue integrable precisely when the integral of its absolute value is finite. For every integer $k\ge 1$, on $I_k=[k+1/6,k+5/6]$ one has $|\sin \pi x|\ge 1/2$ and $x\le k+1$. ConsequentlyThe intervals have disjoint interiors. Countable additivity of the nonnegative Lebesgue integral, or the monotone convergence theorem applied to their finite unions, givesThe harmonic series diverges, so the function is not Lebesgue integrable. Cancellation between adjacent oscillations cannot give absolute integrability.