Results 1 to 2 of 2

Thread: lebesgue measurable function

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    2

    lebesgue measurable function

    Q:
    Prove that if $\displaystyle f : \mathbb{R} \longrightarrow [o,\infty] $ and $\displaystyle f^{-1}((r,\infty]) \in M $ for each $\displaystyle r \in \mathbb{Q} $, then $\displaystyle f $ is lebesgue measurable. ($\displaystyle M $ is the $\displaystyle \sigma $-algebra of Lebesgue measurable sets)

    Attempt:
    I need to show that for any set $\displaystyle A \in \mathcal{B}_{[0,\infty]} $, $\displaystyle f^{-1}(A) \in M $ (is this correct?). So I take a set $\displaystyle A \in \mathcal{B}_{[0,\infty]} $. By definition of $\displaystyle \mathcal{B}_{[0,\infty]} $, $\displaystyle A $ is an open interval (is this correct?). Now, i want to write $\displaystyle A $ as a union of sets of the form $\displaystyle (r,\infty] $ for each $\displaystyle r \in \mathbb{Q} $ but im sturggling to do this! Is this the right method? Can you help me? Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    I need to show that for any set $\displaystyle A \in \mathcal{B}_{[0,\infty]}, f^{-1}(A) \in M$ (is this correct?).
    I think so, because the Lebesgue measure is a measure over the Borelian sigma-algebra of $\displaystyle \mathbb{R}$.

    So I take a set $\displaystyle A \in \mathcal{B}_{[0,\infty]}$. By definition of $\displaystyle \mathcal{B}_{[0,\infty]}$, A is an open interval (is this correct?).
    No, we just know that $\displaystyle \mathcal{B}_{[0,\infty)}$ is generated by the open intervals in $\displaystyle [0,\infty)$
    In other words, it's generated by $\displaystyle \{(a,\infty) \mid a\in \mathbb{R}_+\}$
    But an element of this will not necessarily be an open set.


    It may be useful to use the fact that $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$ (same goes for the positive numbers only).
    Thus, $\displaystyle \forall a \in\mathbb{R}_+$, there exists a (decreasing) sequence $\displaystyle (a_n)_{n\geq 0}$, where $\displaystyle \forall n \in\mathbb{N} ~,~ a_n \in \mathbb{Q}_+$ such that $\displaystyle \lim_{n\uparrow \infty} a_n=a$
    And we can write : $\displaystyle (a,\infty)=\bigcup_{n\geq 0} [a_n,\infty)$

    But one can prove that $\displaystyle f^{-1}\left(\bigcup_{j\in J} A_j\right)=\bigcup_{j\in J} f^{-1}(A_j)$ (where J is a set without any restriction on it). It's one of Haussdorff formulas.

    So you'll have $\displaystyle f^{-1}((a,\infty))\in M$, $\displaystyle \forall a\in \mathbb{R}_+$

    --------------------
    Now the second part (to finish) isn't easier imo.

    There's a lemma that we called "Transport lemma", but I didn't find an English page of it... Here is the French one : Lemme de transport - Wikipédia
    ******transcription******
    X,Y 2 sets. f : X -> Y. $\displaystyle \mathcal{E}\subset \mathcal{P}(Y)$ ( = the power set of Y = the family of all subsets of Y).
    Then $\displaystyle f^{-1}(\sigma(\mathcal{E}))=\sigma(f^{-1}(\mathcal{E}))$

    Here we have $\displaystyle \mathcal{E}=\{(a,\infty) \mid a\in\mathbb{R}_+\}$

    But we know that $\displaystyle f^{-1}(\mathcal{E})\in M$
    Hence, since M is a sigma-algebra, we have the (red) inclusion $\displaystyle f^{-1}(\sigma(\mathcal{E}))=\sigma(f^{-1}(\mathcal{E}))~{\color{red}\subset}~ M$

    And thus for any element A in $\displaystyle \sigma(\mathcal{E})=\mathcal{B}_{[0,\infty)}$, $\displaystyle f^{-1}(A)\in M$



    Whew... Does it look clear ? And I'm pretty sure there's a more elegant way to prove it, but I'm not an expert...
    Well, you have a method, and maybe it will give you guidelines for building your own proof...
    (as you can see, I'm very self-confident ^^)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. E-E for a Lebesgue measurable set E contains some box?
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Nov 21st 2011, 02:26 PM
  2. a question about Lebesgue measurable function
    Posted in the Differential Geometry Forum
    Replies: 15
    Last Post: Jun 29th 2011, 09:53 PM
  3. Non-Lebesgue Measurable
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 4th 2010, 04:11 PM
  4. Lebesgue measurable function and Borel measurable function
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Jan 16th 2010, 02:55 AM
  5. lebesgue measurable set
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Jan 11th 2010, 12:01 AM

Search Tags


/mathhelpforum @mathhelpforum