Past exam of the integration course of the University of Oxford / 2017 / 1 / c

For a Lebesgue measurable function $f:\mathbb{R}\to \mathbb{R}$, the preimage $f^{-1}(U)$ of every open set $U$ is a Lebesgue measurable set. For a Borel measurable function, each such preimage must instead be a Borel set. Equivalently, both definitions can test all Borel target sets: the collectionis a sigma-algebra, since preimages preserve complements and countable unions. Containing all open sets therefore makes it contain the Borel sigma-algebra.

Now let $U$ be open. Since $h$ is a Borel measurable function, $h^{-1}(U)$ is Borel. Since $g$ is a Lebesgue measurable function, $g^{-1}(h^{-1}(U))$ is a Lebesgue measurable set. ButThus $h\circ g$ is a Lebesgue measurable function. The intermediate sigma-algebra is Borel; this is why Borel postcomposition preserves Lebesgue measurability.

The dominated convergence theorem says that if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le H$ almost everywhere for a single Lebesgue integrable function $H\ge 0$, then $f$ is integrable andHere, for $x>0$, put $H_0(x)=x^{-1/2}{e}^{-x}$ and $f_n(x)=H_0(x)(1-(-1{)}^n{e}^{-nx})$. The exponential function gives $e^{-nx}\to 0$ at every such $x$, so $f_n\to H_0$. Also $|f_n(x)|\le 2H_0(x)$. This dominator is integrable: on $(0,1)$, $H_0(x)\le x^{-1/2}$, whose integral is two; on $[1,\infty )$, $H_0(x)\le e^{-x}$, whose integral is finite. DCT therefore proves the required limit.

The limiting value can also be evaluated. Substitute $x=t^2$ and use the Gaussian integral:In fact, scaling the second exponential term gives the exact integral $\sqrt{\pi }[1-(-1{)}^n/\sqrt{n+1}]$. The integral error decays as $n^{-1/2}$ despite exponential convergence at each fixed positive $x$; the slow part lies near the endpoint zero.