Results 1 to 5 of 5

Thread: Measurable Functions

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    63

    Measurable Functions

    Let $\displaystyle (f_n)$ be a sequence of real-valued measurable functions on $\displaystyle [a,b]$. If $\displaystyle (f_n)$ converges to a function $\displaystyle f$ almost everywhere on $\displaystyle [a,b]$, prove that there exists a sequence $\displaystyle (E_n)$ of measurable subsets of $\displaystyle [a,b]$ such that
    $\displaystyle m([a,b] \setminus \cup E_n) = 0$
    and $\displaystyle (f_n)$ converges uniformly to $\displaystyle f$ on each $\displaystyle E_n$.

    How do I even start this question?

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,455
    Thanks
    1368

    Re: Measurable Functions

    If $\displaystyle (f_n)$ converges almost everywhere to $\displaystyle f$, then there is a zero set of points on which it does not converge. A zero set has certain properties. For one thing, it has measure zero. So, if $\displaystyle \[a,b\] \setminus \cup E_n = Z\text{ where }Z\text{ is your zero set}$, then you are guaranteed that the measure will be zero. So, let's construct $\displaystyle E_1$. Since the zero set is a subset of $\displaystyle \[a,b\]$, by the well ordering principle of real numbers, there exists a greatest lower bound of $\displaystyle Z$. Can you find a really small interval $\displaystyle E_1$ that will include $\displaystyle a$, but not $\displaystyle a_1$? Will that interval be measurable? How do you know? Will $\displaystyle (f_n)$ converge to $\displaystyle f$ uniformly on that interval? Again, how do you know?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    63

    Re: Measurable Functions

    Is a_1 a point in the set Z or just part of the partition of [0,1]?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,455
    Thanks
    1368

    Re: Measurable Functions

    It is a point in the set Z.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    Posts
    63

    Re: Measurable Functions

    Alright I managed to solve it. It's actually applying Egorov's Theorem to each E_n and intersect [a,b]\E_n for all n to get m([a,b]\(\cup E_n)) = 0 and f_n converges uniformly to f on each E_n.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Measurable Functions (2)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 26th 2011, 06:39 AM
  2. Measurable functions.
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Apr 27th 2011, 09:55 AM
  3. Measurable Functions
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 18th 2010, 08:37 AM
  4. measurable functions
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 28th 2009, 08:50 AM
  5. measurable functions
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 28th 2008, 09:39 PM

/mathhelpforum @mathhelpforum