Is there a useful relationship between pointwise and $L^2$ distance? It would be really convenient to get a bound on the point-wise closeness of functions by knowing their $L^2$ distance. Clearly, if two functions are close in the $L^2$ sense, you cannot get a general bound on their point-wise distance, but maybe they are close along some parts of their domain.
In my mind a theorem might go something like this:

Given a measurable, bounded function $f:U\rightarrow\mathbb{R}$, if $||f||_2<\varepsilon$ then there is some 'small'  $\delta > 0$ and a set $V\subset U$ where $|f(x)|<\delta \ \forall\  x\in V,$ and $ \mu(V)\geq\mu(U)-\eta_{\varepsilon, \ \delta}$

Then if we had $f_1,f_2\ni||f_1-f_2||_2<\varepsilon$ there would be a 'fairly large' set $V$ where $|f_1-f_2| < \varepsilon$ throughout $V$.
Do theorems like this exist?
 A: The following is a naive take on this problem.
If $f: U \rightarrow \mathbb{R}$ is a measurable function 
such that
$$\sqrt{\int_U f^2} < \epsilon_1$$
and $V \subset U$ is the measurable set such that $\left|f(x)\right| \geq \epsilon_2$ for all $x \in V$ and $\left|f(x)\right| < \epsilon_2$ for all
$x \in U\setminus V$, then
$$\epsilon_2^2 m(V) = \int_{V}\epsilon_2^2 \leq \int_{U}f^2 < \epsilon_1^2$$
Hence 
$$m(V) < \frac{\epsilon_1^2}{\epsilon_2^2}$$
It's not difficult to construct simple functions that show that $m(V)$ can be arbitrarily close to this upper bound. Thus without additional conditions this is the best bound we can get on the size of $V$. 
This suggests the result,
Proposition: Let $\epsilon > 0$. If $f:U \rightarrow \mathbb{R}$ is a measurable function such
that
$$\sqrt{\int_U f^2} < \epsilon$$ 
then the set 
$V = \left\{x \in U : \left|f(x)\right| \geq \sqrt{\epsilon}\right\}$
has measure $m(V) < \epsilon$.
Perhaps with additional conditions a stronger result can be proven.
