Results 1 to 4 of 4

Thread: almost everywhere

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    24

    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?
    Last edited by PRLM; Dec 24th 2009 at 09:06 PM. Reason: fixing a sentence that didnt make sense.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    Of course.
    Notice that {E_n} is an increasing sequence converges to E.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2009
    Posts
    24

    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    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.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum