if $f$ is measurable on $E$ then for all $ \epsilon$ there exist an $F \subset E$ s.t. $ f$ is bounded on all $E$ except $F$ and $m (F) < \epsilon$ This is a question from Royden chapter 3, and this is what my liberal arts mind could think.
I know for sure that it is not elegant but is it completely wrong?


*

*Let there is no such subset for some $\epsilon.$

*Because f is unbounded of  this set $F$,$ f^{-1}F \subset $of all $f^{-1}(a,\infty) \cap E$, and all of these have a measure of more than $\epsilon $  


3.let $f^{-1}(n,\infty)= I_n,$ Then $\lim m (I_n) = \epsilon$, but $ m (\lim I_n) = 0$  hence the contradiction
Thanks,
 A: Here is useful fact about measures of nested sequences:

Suppose
$$ A_1 \supset A_2 \supset A_3 \supset A_4 \supset \dots$$
is a nested sequence of measurable sets, and suppose $m(A_1) < \infty$. Then
$$ \lim_{n \to \infty} m(A_n) = m\left( \cap_{n \in \mathbb N} A_n\right).$$

This statement is very intuitive, and it can be proven rigorously by applying countable additivity of the measure to the disjoint sets $A_1 - A_2$, $A_2 - A_3$, $A_3 - A_4, \dots$
So how does this apply to our question? We are given a measurable function $f : E\to \mathbb R$ with $m(E) < \infty$ (see your comment), and our goal is to prove that for every $\varepsilon > 0$, there exists a measurable $F$ with $m(F) < \varepsilon $ such that $f$ is bounded on $E - F$.
The trick is to define a sequence of subsets $A_n \subset E$:
$$ A_n = f^{-1}((-\infty, -n) \cup (n, \infty)), \ \ \ \ \ n \in \mathbb N$$
These sets obey the property
$$ A_1 \supset A_2 \supset A_3 \supset A_4 \supset  \dots $$
and moreover, they are measurable because the function $f$ is measurable, and $m(A_1) < \infty$ holds because $m(E) < \infty$.
The reason these $A_n$'s are useful is that, for any $n \in \mathbb N$, the function $f$ is guaranteed to be bounded on $E - A_n$. In fact, by the definition of $A_n$, we have $|f(x)| \leq n$ for all $x \in E - A_n$.
So it only remains to show that for any $\varepsilon > 0$, we can find some value of $n$ such that $m(A_n) < \varepsilon$. (If we find such an $n$, we can take $F$ to be $A_n$ for this value of $n$.)
It is enough to show that $\lim_{n \to \infty} m(A_n) = 0$. By our "fact", this is the same thing as showing that $m(\cap_{n \in \mathbb N} A_n) = 0$. And if you think about it, $\cap_{n \in \mathbb N}A_n$ is the empty set, so $m(\cap_{n \in \mathbb N} A_n)$ is certainly zero.
