Past exam of the options I course of the University of Oxford / 2010 / D (Integration)

For every real $b$,Each set on the right is measurable by hypothesis. Closure of a sigma-algebra under countable unions proves the assertion.

For every real $a$,Part (1), applied to measurable $g$, makes the set being complemented measurable. Its relative complement in measurable $E$ is measurable too. The permitted half-line criterion therefore proves $-g$ is measurable.

If $f$ is measurable, then for a threshold $a\le 0$, $\{f^{+}\ge a\}=E$, while for $a>0$, $\{f^{+}\ge a\}=\{f\ge a\}$. The half-line criterion makes $f^{+}$ measurable. By part (2), $-f$ is measurable, and $f^{-}=(-f{)}^{+}$ is therefore measurable as well.

Conversely, if $f^{+}$ and $f^{-}$ are measurable, part (2) and the permitted closure under sums show that $f=f^{+}-f^{-}$ is measurable. Thus measurability is equivalent to measurability of both parts.

Since $|f|=f^{+}+f^{-}$, part (3) and closure under sums prove that $|f|$ is measurable whenever $f$ is.

The converse is false. Let $A\subseteq [0,1]$ be a non-Lebesgue-measurable set and define $f=1$ on $A$ and $f=-1$ on its complement in $[0,1]$. Then $|f|=1$ is measurable, but $f^{-1}([0,\infty ))=A$ is not. Taking absolute values can conceal a nonmeasurable sign choice.

Both functions are integrable on $(0,1)$. The oscillatory function is measurable and bounded by $1$ in absolute value, on an interval of finite measure. For the other function,In particular its singularity at the right endpoint is of integrable inverse-square-root type.

For $x\ge 1$, $|\sin (1/\sqrt{x})|\le 1/\sqrt{x}$, soThus $g$ is integrable on this interval. But for $x>1$, $\sqrt{x^2-1}<x$, hence $h(x)>1/x$. The integral of $1/x$ diverges at infinity, so $h$ is not integrable. Defining $h(1)=0$ changes only a null set and has no effect on this conclusion.