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Given a measure space $\Omega$.

Consider a sequence of measurable functions $f_n$

Suppose it converges pointwise: $f_n\to f$

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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?

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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 $\int_{A_{N}}\left|f_{n}-f\right|\,{\rm d}\mu\to0$.

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.

$$

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