Very silly question about an inequality in integration of two functions I have a very silly question about an inequality.
$$\left|\int (f-g)\right| \leq \int |f-g|$$
I know this is true in general, but I am not sure how to prove it to myself, I know it does not depend on of the type of integration as long as it is additive. I am currently doing it in the sense of Lebesgue and bounded functions. Can someone help me see why it is true?
 A: Notice you only need to show it for $|\int f | \le \int |f|$. Now $-|f|\le f \le |f|$. Can you finish from there?
A: This is a special case of the general principle that $|\int f|\le \int |f|$ with $f-g$ in the place of $f$.  This is because of the definition of the lebesgue integral,  as $\int f=\int f^+ -\int f^-$,  where $f^+$ is the positive part of f, and $f^-$ is the positive part of $-f$.  So we have $|f|\ge |f^+-f^-|$, and we are done.
A: Here is a proof for complex valued functions.  Suppose $f : X \to \Bbb C$.  We want to show $\left |\int f \,d\mu \right | \leq  \int |f| \,d\mu $.
Note that since $ \int f \,d\mu$ is a complex number, if $\overline{\int f \,d\mu}$ denotes its conjugate, we have that:
$\left | \int f \,d\mu \right | = \dfrac{\left | \int f \,d\mu \right |^{2}}{\left | \int f \,d\mu \right |} =\dfrac{\overline{ \int f \,d\mu }\cdot\int f \,d\mu}{| \int f \,d\mu|}$
Now if we let $z = \dfrac{\overline{ \int f \,d\mu }}{|\int f \,d\mu|}$, then $z$ is a complex number with $|z| = 1$.
So, we have $|\int f \,d\mu| = \int z f \,d\mu$.  Since the left hand side is a real number, it means the imaginary part of $\int zf \,d\mu$ is $0$, so we have:
$|\int f \,d\mu| = \int zf \,d\mu = Re(\int zf \,d\mu) =  \int Re(zf) \,d\mu \leq \int |Re(zf)|\,d\mu \leq \int |f| \,d\mu$
We got the last inequality since $(Re(zf))^{2} \leq (Re(zf))^{2} + (Im(zf))^{2} = |f|^{2}$, so $|Re(zf)| \leq |f|$ by taking the square root of both sides of the inequality.
