# Measure Spaces: Uniform & Integral Convergence

Given a measure space $\Omega$.

Consider a sequence of measurable functions $f_n$

Suppose it converges pointwise: $f_n\to f$

Can one find increasing subsets with uniform convergence: $$A_N\uparrow\Omega:\quad\|f-f_n\|_{A_N}\stackrel{n\to\infty}{\to}0$$ Or at least with convergence in integral norm: $$A_N\uparrow\Omega:\quad\int_{A_N}|f-f_n|\mathrm{d}\mu\stackrel{n\to\infty}{\to}0$$ (Not passing over to subsequences!)

Some first basic examples allow that: $$f_n:=\chi_{(n,n+1]}:\quad\|f-f_n\|_{A_N}\stackrel{n\to\infty}{\to}0\quad(A_N:=(0,N])$$ $$g_n:=x^n:\quad\|g-g_n\|_{B_N}\stackrel{n\to\infty}{\to}0\quad(B_N:=(0,1-\frac{1}{N}])$$ $$h_n:=x^{1/n}:\quad\|h-h_n\|_{C_N}\stackrel{n\to\infty}{\to}0\quad(C_N:=(\frac{1}{N},1])$$

So I wonder: Are there are examples where both can't hold?

• First of all, note that the first statement is not really a measure theoretic claim. If it is true, we can just use the whole power set as the class of measurable sets, so that measurability is always satisfied. Hence, this is a (still interesting) statement about pointwise convergence per se (and hence probably false in general). For the integral question, one can certainly construct a measure $\mu$ (even Lebesgue measure) and a sequence such that $\int_A |f_n -f|\, d\mu = \infty$ for each set $A$ of positive measure and all $n$, but still $f_n \to f$ pointwise. – PhoemueX Dec 14 '14 at 20:45
• @PhoemueX: Ah right, so one could just take aaany measure on the power set; the statement itself is absolutely independent of measures. – C-Star-W-Star Dec 14 '14 at 20:50
• @PhoemueX: Do you have a counterexample for the second statement; for convenience say $\Omega:=[0,1]$ with Lebesgue measure $\mu:=\lambda$ and w.l.og. $f_n\to0$? – C-Star-W-Star Dec 14 '14 at 21:16
• For a space of finite measure, there can be no counterexample, due to Egorov's theorem (en.wikipedia.org/wiki/Egorov%27s_theorem). Take $\Omega_n \subset \Omega$ with $\Omega_n \subset \Omega_{n+1}$ and $\mu(\Omega \setminus \Omega_n) < 1/n$ and $f_N \to f$ uniformly on $\Omega_n$ and then let $A_n := \Omega_n \cup [\Omega \setminus \bigcup_n \Omega_n]$, where you should observe that the set in brackets is a null-set. I will this (together with a counterexample on $\Bbb{R}$ with Lebesgue measure) tomorrow. ATM I do not have the time :( – PhoemueX Dec 14 '14 at 21:19
• @PhoemueX: Ok and for $\Omega:=\mathbb{R}$ with Lebesgue measure $\mu:=\lambda$ and w.l.o.g. $f_n\to0$? – C-Star-W-Star Dec 14 '14 at 21:21

The first part of the question (about uniform convergence on the $A_{N}$) turns out to be false. To see this, let us consider the set $$\Omega:=\left\{ \left(x_{n}\right)_{n\in\mathbb{N}}\in c_{0}\left(\mathbb{N}\right)\,\mid\, x_{n}\geq0\text{ for all }n\in\mathbb{N}\right\}$$ of all non-negative null-sequences. It is easy to see that this is a closed subset of the Banach space $c_{0}\left(\mathbb{N}\right)$ of null-sequences (considered itself as a (closed) subspace of $\ell^{\infty}\left(\mathbb{N}\right)$). Hence, $\Omega$ is complete. As $c_{0}\left(\mathbb{N}\right)$is separable, $\Omega$ is even a Polish space. We equip $\Omega$ with the Borel $\sigma$-Algebra (but taking the whole power set would also be fine).

Let us now define $$f_{n}:\Omega\to\mathbb{R}_{+},\left(x_{m}\right)_{m\in\mathbb{N}}\mapsto x_{n}.$$ Then each of the $f_{n}$ is continuous, hence measurable. Furthermore, for $\left(x_{m}\right)_{m\in\mathbb{N}}\in\Omega\subset c_{0}\left(\mathbb{N}\right)$ arbitrary, we have $$f_{n}\left(\left(x_{m}\right)_{m\in\mathbb{N}}\right)=x_{n}\xrightarrow[n\to\infty]{}0$$ and hence $f_{n}\to0$ pointwise.

Let us now assume that there is some sequence $\left(A_{N}\right)_{N\in\mathbb{N}}$ of subsets $A_{N}\subset\Omega$ with $$\left\Vert f_{n}\right\Vert _{A_{N},\infty}:=\sup_{x\in A_{N}}\left|f_{n}\left(x\right)\right|\xrightarrow[n\to\infty]{}0,$$ i.e. the convergence is uniform on each of the $A_{N}$ and $\Omega=\bigcup_{N\in\mathbb{N}}A_{N}$. Using the continuity of the $f_{n}$, we get $\left\Vert f_{n}\right\Vert _{A_{N},\infty}=\left\Vert f_{n}\right\Vert _{\overline{A_{N}},\infty}$, where $\overline{A_{N}}$ is the closure of $A_{N}$ (in $\Omega$). Hence, we can assume w.l.o.g. that the $A_{N}$ are closed.

By Baire's category theorem, there is some $N\in\mathbb{N}$ and some $x=\left(x_{m}\right)_{m\in\mathbb{N}}\in A_{N}\subset\Omega$ such that $x$ is an interior point of $A_{N}$ (considered as a subset of $\Omega$), i.e. there is some $r>0$ with $$\left(\Omega\cap B_{r}\left(x\right)\right)\subset A_{N},$$ where the ball $B_{r}\left(x\right)$ is formed in $c_{0}$ (or $\ell^{\infty}$, this does not matter, because we intersect with $\Omega\subset c_{0}\subset\ell^{\infty}$).

Regard the Kroneckers: $(0,\ldots,0,1,0,\ldots)\in\Omega$

For arbitrary $n\in\mathbb{N}$, this implies $x+\frac{r}{2}\delta_{n}\in\Omega\cap B_{r}\left(x\right)\subset A_{N}$ and hence $$\frac{r}{2}\leq x_{n}+\frac{r}{2}=f_{n}\left(x+\frac{r}{2}\delta_{n}\right)\leq\left\Vert f_{n}\right\Vert _{A_{N},\infty}\xrightarrow[n\to\infty]{}0,$$ a contradiction.

By standard results on Polish spaces, any two uncountable Polish spaces are Borel-isomorphic. This shows that the result also fails on $\mathbb{R}$ equipped with the Borel $\sigma$-Algebra.

For completeness, let me repeat the results for the second part of the question (regarding $\int_{A_{N}}\left|f_{n}-f\right|\,{\rm d}\mu\to0$) given in the comments.

If $\mu\left(\Omega\right)$ is a finite measure, Egoroff's theorem (http://en.wikipedia.org/wiki/Egorov\%27s_theorem , one of my favourite theorems) gives for each $M\in\mathbb{N}$ some measurable subset $B_{N}\subset\Omega$ such that $\mu\left(\Omega\setminus B_{N}\right)<\frac{1}{N}$ and such that $f_{n}|_{B_{N}}\to f|_{B_{N}}$ uniformly. Then $$M:=\Omega\setminus\bigcup_{N\in\mathbb{N}}B_{N}=\bigcap_{N\in\mathbb{N}}\left(\Omega\setminus B_{N}\right)$$ is a $\mu$-null-set, so that $$\int_{B_{N}\cup M}\left|f_{n}-f\right|\,{\rm d}\mu=\int_{B_{N}}\left|f_{n}-f\right|\,{\rm d}\mu\leq\mu\left(B_{N}\right)\cdot\left\Vert f_{n}-f\right\Vert _{\infty,B_{N}}\xrightarrow[n\to\infty]{}0.$$ Now set $A_{N}:=M\cup B_{1}\cup\dots\cup B_{N}$. This yields $A_{N}\uparrow\Omega$ and $$\int_{A_{N}}\left|f_{n}-f\right|\,{\rm d}\mu\leq\sum_{\ell=1}^{N}\int_{M\cup B_{\ell}}\left|f_{n}-f\right|\,{\rm d}\mu\xrightarrow[n\to\infty]{}0.$$

This even generalizes to the $\sigma$-finite case: If $\Omega=\biguplus_{m\in\mathbb{N}}\Omega_{m}$, with each $\Omega_{m}$ of finite measure, the preceding argument yields sets $\Omega_{m,N}$ with $\Omega_{m,N}\uparrow\Omega_{m}$ and $\int_{\Omega_{m,N}}\left|f_{n}-f\right|\,{\rm d}\mu\to0$. Now take any surjection $\pi:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$ and let $$B_{N}:=\Omega_{\left(\pi\left(N\right)\right)_{1},\left(\pi\left(N\right)\right)_{2}}$$ as well as $A_{N}:=B_{1}\cup\dots\cup B_{N}$. This yields $A_{N}\uparrow\Omega$ and $\int_{B_{N}}\left|f_{n}-f\right|\,{\rm d}\mu\to0$ for each $N$, which (as above) also yields $\int_{A_{N}}\left|f_{n}-f\right|\,{\rm d}\mu\to0$.

## One of the remarkable facts here is that we did not assume $f_n$ or $f$ to be integrable to begin with!

In the non-$\sigma$-finite case, this does not remain true any longer. We can for example take the counting measure $\mu:=\#$ on $\Bbb{R}$ and $f_n \equiv 1/n$. If $\Bbb{R} = \bigcup_N A_N$, then at least one of the sets $A_N$ is uncountable, which yields

$$\int_{A_N} |f_n - f | \, d\mu = \frac{1}{n} \cdot \#(A_N) = \infty \not \to 0.$$

• @Freeze_S: As always, feel free to edit if you feel the need. – PhoemueX Dec 15 '14 at 21:58
• Nice answer, really!! :) (For the sigma finite case one could as well take from the beginning $\Omega_m\uparrow\Omega$, that would save some lines but in principle it is the same.) – C-Star-W-Star Dec 16 '14 at 1:56