I am struggling with this exercise:
Suppose that $f \in L^1(\Omega, A, \mu)$. Show that for each $\epsilon > 0$, there is a $\delta >0$ such that $\mu(E) < \delta \rightarrow\int_E|f| \mu<\epsilon$.
Since I know that f is integrable, I know that $\int_\Omega|f|d\mu=K<\infty$. If this K is zero, I see that I allready have what I want, so I can assume that K is bigger than zero.
I am having a problem since I don't know if the function is bounded, had the function been bounded by a number M, I could just choose $\delta=\epsilon/(2\cdot M)$.
I think the $\delta$ should be a function of K and $\epsilon$ somehow.
Any tips?
Note that the statement would be trivial if the function $f$ were bounded.
So set $f_n(x) = f(x)$ if $f(x) \leq n$ and $f(x) = n$ otherwise. Then each $f_n$ is bounded and converges to $f$ point wise. By the monotone convergence theorem, given $\epsilon > 0$ there exists $N$ such that $\int_E f - \int_E f_N < \epsilon/2$. So choose $\delta < \epsilon/(2N)$. If $\mu(A) < \delta$, we have that
$\int_A f = \int_A (f - f_N) + \int_A f_N < \epsilon/2 + \epsilon/2$ as needed.