The integral over a subset is smaller? In a previous question I had $A \subset \bigcup_{k=1}^\infty R_k$ where $R_k$ in $\Bbb{R}^n$ are rectangles I then proceeded to use the following inequality $\left|\int_A f\right| \le \left|\int_{\bigcup_{k=1}^\infty R_k} f \right|$ which I am not really certain of. Does anyone know how to prove it? If its wrong what similar inequality should I use to prove the result here.

For $A \subset B$, $$
  \left|\int_A f\right| \leq \left|\int_B f \right|
$$
is true if either $f(x) \geq 0$ for all $x \in B$, or $f(x) \leq 0$ for all $x \in B$.
For a counter-example if this isn't the case, let $f(x) = x$, $A = [0,2]$, $B=[-2,2]$. Then $$
   \left|\int_A f\right| = 2 > 0 = \left|\int_B f \right| \text{.}
$$
In particular, you thus always have that $$
  \int_A |f| \leq \int_B |f| \
$$
if $A \subset B$.
