In my homework, I have some trouble with the following problem in measure space $(X,\mathcal{A},\mu)$:
Let $K_n=\{|u| \leq n\}$
Show that:
$|\int_E u d\mu|\leq n\cdot \mu(E)+\int_{X\backslash K_n} |u| d\mu$
for all $n\geq 1$ and all $E \in \mathcal{A}$
I have gotten so far that I have:
$|\int_E u d\mu|\leq \int_E n d\mu+\int_{X\backslash K_n} |u| d\mu$
I see that when n inceases then the last term of the rightside of the inequality goes to zero and the first term of the rightside increases.
I just can't see the connection between then so that they are always greater than the leftside of the inequality. Any hint/help would be very appreciated
Since $$ \int_E u \, d\mu = \int_{E\cap K_n} u \, d\mu + \int_{E\setminus K_n} u\, d\mu, $$ you have that $$ \left|\int_E u\, d\mu\right| \leq \int_{E\cap K_n} |u| \, d\mu + \int_{E\setminus K_n} |u|\, d\mu \leq \int_{E\cap K_n} n \, d\mu + \int_{E\setminus K_n} |u|\, d\mu\,. $$ Since $E\cap K_n \subset E$, $E\setminus K_n\subset X\setminus K_n$, and $|u| \geq 0$, you finally get the required inequality: $$ \left|\int_E u\, d\mu\right| \leq n \, \mu(E) + \int_{X\setminus K_n} |u|\, d\mu\,. $$