Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 22F (Analysis of Functions)

The pointwise limit $f$ is a measurable function, since it can be expressed as ${\sup }_n{\inf }_{m\ge n}{f}_m$. Thus each set $E_n^{(k)}$ is a countable intersection of measurable sets within $A$. Deleting the condition at index $m=n$ weakens the requirements, whereas replacing $1/k$ by $1/(k+1)$ strengthens them. HenceThese are inclusions, not necessarily strict inclusions. The displayed sets describe eventual uniform error tolerances at an individual point.

Fix $k\ge 1$ and $x\in A$. By pointwise convergence, there is an $N=N(x,k)$ such that $|f_m(x)-f(x)|\le 1/k$ for every $m\ge N$. This is exactly $x\in E_N^{(k)}$, provingTogether with monotonicity in $n$ and finite measure of $A$, continuity from below of a measure gives $\lambda (E_n^{(k)})\uparrow \lambda (A)$. ConsequentlyThis quantitative form is what allows a single large-measure set to work for every tolerance, rather than choosing a different exceptional set for each point.

For each $k$, choose $n_k$ with $\lambda (A\setminus E_{n_k}^{(k)})\le \epsilon 2^{-k-1}$ and set $B={\bigcap }_{k\ge 1}{E}_{n_k}^{(k)}$. Countable subadditivity gives $\lambda (A\setminus B)\le \epsilon /2$. For any tolerance $\delta >0$, choose $k$ with $1/k<\delta$. For all $m\ge n_k$ and every $x\in B$, $|f_m(x)-f(x)|\le 1/k$, proving uniform convergence on $B$.

If the functions are Borel measurable, $B$ itself is Borel and we can take $A_{\epsilon }=B$. If “measurable” is interpreted in the completed Lebesgue sense, use regularity of Lebesgue measure to choose a Borel subset $A_{\epsilon }\subseteq B$ differing from $B$ by a null set; the same measure bound and uniform convergence hold. Thus the requested Borel conclusion is valid under either convention. This proves the finite-measure Egorov theorem.

For an infinite-measure counterexample take $A=[0,\infty )$ and $f_n=1_{[n,\infty )}$. It converges pointwise to zero. Any subset of $A$ with complement of finite measure intersects every infinite tail $[n,\infty )$, so the supremum of $f_n$ there is one for every $n$. Hence finite measure is essential to the conclusion.

For $1<p<\infty$, apply Fatou lemma to the nonnegative sequence $|f_n{|}^p$, which converges pointwise to $|f{|}^p$:For $p=\infty$, each inequality $\Vert f_n{\Vert }_{\infty }\le M$ fails pointwise only on a null set. Outside the union of these countably many null sets, $|f_n(x)|\le M$ for every $n$, so the pointwise limit also satisfies $|f(x)|\le M$. Therefore in both cases

Let $q$ be conjugate to $p$, so $1/q=1-1/p$, including $q=1$ for $p=\infty$. Hölder's inequality implies, for every measurable finite-measure set $E$,This proves integrability on $A$ and gives uniform integrability there. Choose a set $B\subseteq A$ with $\lambda (A\setminus B)\le \delta$ on which convergence is uniform, using Egorov theorem. ThenFirst take $n\to \infty$ and then $\delta \downarrow 0$. Thus even the local $L^1$ difference tends to zero, and in particularFor $1<p<\infty$, simple functions supported on finite-measure Borel sets are dense in $L^q(\mathbb{R})$. The displayed convergence proves pairing convergence for such functions. Approximate any $h\in L^q$ by a simple $h_0$ and useBy the duality of Lp spaces, these are all continuous linear functionals on $L^p$, so $f_n$ converges weakly to $f$. It need not converge strongly: $f_n=1_{[n,n+1]}$ converges pointwise to zero but has $L^p$ norm one. For $p=\infty$ the pairing argument gives weak-star convergence against $L^1$, rather than asserting weak convergence against every functional in the much larger full dual.