Originally Posted by

**Moo** Hello,

Here is a solution (not from me)

Let's fix $\displaystyle \epsilon$

Let $\displaystyle A_n=\left\{ x\mid\left\vert f(x)-f_{n}(x)\right\vert \geq\epsilon\right\}$

Because we need n in it...

We know that $\displaystyle \lim_n \mu(A_n)=0$.

We have $\displaystyle |f|\leq |f_n|+\epsilon$, except over $\displaystyle A_n$ (this deals with inclusion of sets), which is a set which measure goes to 0 as n goes to infinity.

By taking the limit, we can say that $\displaystyle |f|\leq |g|+\epsilon$, except over a set of measure 0.

Thus it is true almost everywhere.

Since this is true for any $\displaystyle \epsilon$, we can take $\displaystyle \epsilon=1/p$ and consider that $\displaystyle \mu\left(\bigcap_{p\geq 1} \{|f|\leq |g|+\tfrac 1p\}\right)=\mu(E)$

But the measure of this intersection is the measure of $\displaystyle \{|f|\leq |g|\}$

Hence $\displaystyle |f|\leq |g|$ almost everywhere.

And finally, we can use Lebesgue's dominated convergence theorem.