Absolute value of integral equal to integral of absolute value

Question

I tried to prove this, however I'm not sure if I did it right. Suppose $$\left|\int f d\lambda\right|=\int|f| d\lambda$$ then $f\leq 0$ or $f\geq0$

My proof: Suppose $f < 0$ then $f=-|f|$, so $$-\int fd\lambda=\int |f| d\lambda=\left|\int fd\lambda\right|$$ Since those numbers are positive, then $-f \geq 0$ a.e.

Is it right and do I have to suppose that $f>0$?

Answer 1

You are proving the converse of what you are asked to prove.

If $\int f d\lambda \geq 0$ then we get $\int (|f|-f)d\lambda=0$. But $|f|-f \geq 0$. If the integral of a non-negative measurable function is $0$ then the function is $0$ a.e. Hence $|f|-f =0$ a.e. so $f=|f| \geq 0$ a.e. When $\int f d\lambda < 0$ the argument is similar. Consider $|f|+f$ instead of $|f|-f$.