Measure Spaces: Uniform & Integral Convergence

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

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