# Thread: almost everywhere

1. ## almost everywhere

Let $\displaystyle f$ be nonnegative measurable function. Show $\displaystyle \int f = 0$ implies $\displaystyle f=0$ almost everywhere.

let $\displaystyle E=\{x:f>0\}$. Then $\displaystyle E=\bigcup E_n$ where $\displaystyle E_n=\{x:f>1/n \}$.
now 0=$\displaystyle \int f \geq \int f1_{E_n} \geq \frac{1}{n}\int 1_{E_n} = \frac{1}{n} mE_n$.
for each $\displaystyle n$, $\displaystyle 1/n \neq 0$ so $\displaystyle mE_n=0$.

Here is my question. i am not sure if i can say $\displaystyle lim_{n \rightarrow \infty} mE_n = 0$. can i say that?

2. Of course.
Notice that {E_n} is an increasing sequence converges to E.

3. ## a.e

i can see that for each n $\displaystyle mE_n=0$ and i guess it makes sense that $\displaystyle lim_{n \rightarrow \infty} mE_n=0$. but i dont know how to show that. would u get me started on it?

4. Since the sequence $\displaystyle \{E_n\}\text{ is an increasing sequence converging to }E,mE_n=0, \forall n$.we obtain
$\displaystyle mE=lim_{n \to \infty} mE_n=lim_{n\to\infty}0=0$
that is what we want.