Results 1 to 5 of 5

Math Help - Null Sets

  1. #1
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347

    Null Sets

    Hello all, I am trying to prove that a countable union of null sets is null, i.e.

    \displaystyle\bigcup_{n=1}^{\infty} X_n is null when each X_n is null. Now I am a bit confused about the notion of an infinite union. Is it enough to show that \displaystyle\bigcup_{n=1}^M X_n is null for all M \in \mathbb{N} ?

    I have managed to do that in the following way: since you pick \epsilon, then for X_1,\ldots X_M, these are all null so you can define sequences \{ I_{n_r} \} with \displaystyle\sum_{r=0}^{\infty}m(I_{n_r}) < \frac{\epsilon}{M+1}. Hence we define a sequence \{ I_r\} by putting all the sequence elements of \{I_{1_r}\} up to \{ I_{M_r}\} in the sequence, ie the sequence \{ I_{1_1}, I_{2_1},\ldots I_{M_1}, I_{2_1},I_{2_2},\ldots I_{M_2}, \ldots \}. Of course \displaystyle\bigcup_{n=1}^{M} X_n \subseteq \displaystyle\bigcup_{r=1}^{\infty} I_r, and we have that \displaystyle\sum_{n=1}^{\infty}m(I_r) < M\frac{\epsilon}{M+1} < \epsilon. Therefore \display\bigcup_{n=1}^{M} X_n is null for all M\in\mathbb{N}.

    But I am confused about the infinite union? Can I deduce it from the above?

    what about choosing \epsilon, then for each X_n, choose a sequence \{ I_{n_r}\} such that \displaystyle\sum_{r=1}^{\infty} m(I_{n_r}) < \frac{\epsilon}{2^{n+1}}. Then I would have to define a sequence \{ I_r\} which somehow contains infinitely many of the above sequences, so that I get \displaystyle\sum_{r=1}^{\infty} I_r < \epsilon. But can I construct a sequence like that?

    Any help would be appreciated, thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    Are you dealing with sets in a measure space? If so, then you can use the property that the measure, call it \mu, is countably subadditive. That is, if \{E_n\} is a countable sequence of sets in your \sigma-algebra, then

    \dipslaystyle \mu(\bigcup E_n)\leq \sum \mu(E_n)

    In particular, if all of the sets E_n are null sets, then the sum on the right is zero, which proves your result.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347
    Thanks for your response. Everything is in  \mathbb{R}. We just started the course and haven't define measure yet, so we're expected to do it from first principles at the moment, but no doubt we will study that which you mentioned at some point. Although I'm not too sure about this, the course description says that we will define the lebesgue integral, and measure theory will follow from a corollary.

    I feel a bit more confident about my solution. Indeed I can define a sequence just by writing the infinite list of sequences in a table starting at the top, writing all the elements of the first sequence in a row, then in the next row write the elements of the second sequence, and so on, and this is of course countable by the same reason \mathbb{Q} is, by drawing a line that "snakes" from the top left of the table.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,398
    Thanks
    1848
    Quote Originally Posted by slevvio View Post
    Hello all, I am trying to prove that a countable union of null sets is null, i.e.

    \displaystyle\bigcup_{n=1}^{\infty} X_n is null when each X_n is null. Now I am a bit confused about the notion of an infinite union. Is it enough to show that \displaystyle\bigcup_{n=1}^M X_n is null for all M \in \mathbb{N} ?
    No, you can't. That would show that the union of any finite collection of null sets is null. No matter how large N is, it is still finite.

    I have managed to do that in the following way: since you pick \epsilon, then for X_1,\ldots X_M, these are all null so you can define sequences \{ I_{n_r} \} with \displaystyle\sum_{r=0}^{\infty}m(I_{n_r}) < \frac{\epsilon}{M+1}. Hence we define a sequence \{ I_r\} by putting all the sequence elements of \{I_{1_r}\} up to \{ I_{M_r}\} in the sequence, ie the sequence \{ I_{1_1}, I_{2_1},\ldots I_{M_1}, I_{2_1},I_{2_2},\ldots I_{M_2}, \ldots \}. Of course \displaystyle\bigcup_{n=1}^{M} X_n \subseteq \displaystyle\bigcup_{r=1}^{\infty} I_r, and we have that \displaystyle\sum_{n=1}^{\infty}m(I_r) < M\frac{\epsilon}{M+1} < \epsilon. Therefore \display\bigcup_{n=1}^{M} X_n is null for all M\in\mathbb{N}.

    But I am confused about the infinite union? Can I deduce it from the above?

    what about choosing \epsilon, then for each X_n, choose a sequence \{ I_{n_r}\} such that \displaystyle\sum_{r=1}^{\infty} m(I_{n_r}) < \frac{\epsilon}{2^{n+1}}. Then I would have to define a sequence \{ I_r\} which somehow contains infinitely many of the above sequences, so that I get \displaystyle\sum_{r=1}^{\infty} I_r < \epsilon. But can I construct a sequence like that?

    Any help would be appreciated, thank you
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by slevvio View Post
    Hello all, I am trying to prove that a countable union of null sets is null, i.e.

    \displaystyle\bigcup_{n=1}^{\infty} X_n is null when each X_n is null. Now I am a bit confused about the notion of an infinite union. Is it enough to show that \displaystyle\bigcup_{n=1}^M X_n is null for all M \in \mathbb{N} ?

    I have managed to do that in the following way: since you pick \epsilon, then for X_1,\ldots X_M, these are all null so you can define sequences \{ I_{n_r} \} with \displaystyle\sum_{r=0}^{\infty}m(I_{n_r}) < \frac{\epsilon}{M+1}. Hence we define a sequence \{ I_r\} by putting all the sequence elements of \{I_{1_r}\} up to \{ I_{M_r}\} in the sequence, ie the sequence \{ I_{1_1}, I_{2_1},\ldots I_{M_1}, I_{2_1},I_{2_2},\ldots I_{M_2}, \ldots \}. Of course \displaystyle\bigcup_{n=1}^{M} X_n \subseteq \displaystyle\bigcup_{r=1}^{\infty} I_r, and we have that \displaystyle\sum_{n=1}^{\infty}m(I_r) < M\frac{\epsilon}{M+1} < \epsilon. Therefore \display\bigcup_{n=1}^{M} X_n is null for all M\in\mathbb{N}.

    But I am confused about the infinite union? Can I deduce it from the above?

    what about choosing \epsilon, then for each X_n, choose a sequence \{ I_{n_r}\} such that \displaystyle\sum_{r=1}^{\infty} m(I_{n_r}) < \frac{\epsilon}{2^{n+1}}. Then I would have to define a sequence \{ I_r\} which somehow contains infinitely many of the above sequences, so that I get \displaystyle\sum_{r=1}^{\infty} I_r < \epsilon. But can I construct a sequence like that?

    Any help would be appreciated, thank you
    As HallsofIvy pointed out it looks as though you did it for arbitrarily finite number of sets, but I think what you're saying makes sense. Let \varepsilon>0 be given. Since each X_n we can choose an infinite cover \displaystyle \Omega_n=\{I_{r,n}\}_{r\in\mathbb{N}} of it such that \displaystyle \sum_{r=1}^{\infty}\mu\left(I_r\right)<\frac{\vare  psilon}{2^{n+1}}. So, let \displaystyle \Omega=\bigcup_{n=1}^{\infty}\Omega_n=\left\{I_m\}  _{m\in\mathbb{N}}. Note then that it shouldn't be too hard (recalling countable subadditivity) to prove that \displaystyle \sum_{m=1}^{\infty}\mu\left(I_m\right)\leqslant \sum_{n=1}^{\infty}\sum_{r=1}^{\infty}\mu\left(I_{  n,r}\right)=\sum_{n=1}^{\infty}\frac{\varepsilon}{  2^{n+1}}=\varepsilon
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. null
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: December 16th 2013, 03:51 PM
  2. Metric spaces, open sets, and closed sets
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: March 16th 2011, 06:17 PM
  3. Replies: 9
    Last Post: November 6th 2010, 01:47 PM
  4. Null Sets and uniform convergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 16th 2010, 03:21 AM
  5. Riemann integral and null sets
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: January 30th 2010, 04:45 PM

Search Tags


/mathhelpforum @mathhelpforum