1. ## 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

2. 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.

3. 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.

4. Originally Posted by slevvio
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

5. Originally Posted by slevvio
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 $\displaystyle \sum_{m=1}^{\infty}\mu\left(I_m\right)\leqslant \sum_{n=1}^{\infty}\sum_{r=1}^{\infty}\mu\left(I_{ n,r}\right)=\sum_{n=1}^{\infty}\frac{\varepsilon}{ 2^{n+1}}=\varepsilon$