# Thread: almost everywhere

1. ## almost everywhere

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

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

Here is my question. i am not sure if i can say $lim_{n \rightarrow \infty} mE_n = 0$. 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 $mE_n=0$ and i guess it makes sense that $lim_{n \rightarrow \infty} mE_n=0$. but i dont know how to show that. would u get me started on it?

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