Proving some properties on a complete measure space

[mahler1]

Homework Statement .

A space $(X,\Sigma, \mu)$ is a complete measure space if given $Z \in \Sigma$ such that $\mu(Z)=0$, for every $Y \subset Z$, we have $Y \in \Sigma$. In this case, prove that

a) If $Z_1 \in \Sigma$, $Z_1ΔZ_2 \in \Sigma$ and $\mu(Z_1ΔZ_2)=0$, then $Z_2 \in \Sigma$.

b) If $E_1, E_2 \in \Sigma$ and $\mu(E_1ΔE_2)=0$, then $\mu(E_1)=\mu(E_2)$.



The attempt at a solution.

For a), $Z_2\setminus Z_1 \subset Z_1ΔZ_2$, by definition of complete measure space, we have that $Z_2\setminus Z_1 \in \Sigma$. If I could prove that $Z_1\cap Z_2 \in \Sigma$, then, I could conclude that $Z_2=(Z_2\setminus Z_1) \cup (Z_1 \cap Z_2) \in \Sigma$. I got stuck trying to prove the last part.

For b), I think I got it but I would like to check if my solution is correct:

First notice that since we are in a measure space and $E_1, E_2 \in \Sigma \implies {E_1}^c, {E_2}^c \in \Sigma \implies E_1 \cap E_2=({E_1}^c \cup {E_2}^c)^c \in \Sigma$. Moreover, since it is a complete measure space, $E_1\setminus E_2, E_2 \setminus E_1 \in \Sigma$. Having said that, we can write $E_1=(E_1 \setminus E_2) \cup (E_1 \cap E_2)$, the same goes for $E_2$. As $\mu$ is a measure, we have that $\mu(E_1)=\mu(E_1 \setminus E_2) + \mu(E_1 \cap E_2)$, and $\mu(E_2)=\mu(E_2 \setminus E_1) + \mu(E_1 \cap E_2)$.

Again, since $\mu$ is a measure, $0\leq \mu(E_1 \setminus E_2),\mu(E_2 \setminus E_1)\leq \mu(E_1ΔE_2)=0 \implies \mu(E_1 \setminus E_2)=0=\mu(E_2 \setminus E_1)$.

But then, $\mu(E_1)=0+\mu(E_1 \cap E_2)=\mu(E_2)$, which was what we wanted to prove.


I would appreciate some help for part a)

 

[Dick]

Homework Statement .

A space $(X,\Sigma, \mu)$ is a complete measure space if given $Z \in \Sigma$ such that $\mu(Z)=0$, for every $Y \subset Z$, we have $Y \in \Sigma$. In this case, prove that

a) If $Z_1 \in \Sigma$, $Z_1ΔZ_2 \in \Sigma$ and $\mu(Z_1ΔZ_2)=0$, then $Z_2 \in \Sigma$.

b) If $E_1, E_2 \in \Sigma$ and $\mu(E_1ΔE_2)=0$, then $\mu(E_1)=\mu(E_2)$.
The attempt at a solution.

For a), $Z_2\setminus Z_1 \subset Z_1ΔZ_2$, by definition of complete measure space, we have that $Z_2\setminus Z_1 \in \Sigma$. If I could prove that $Z_1\cap Z_2 \in \Sigma$, then, I could conclude that $Z_2=(Z_2\setminus Z_1) \cup (Z_1 \cap Z_2) \in \Sigma$. I got stuck trying to prove the last part.

For b), I think I got it but I would like to check if my solution is correct:

First notice that since we are in a measure space and $E_1, E_2 \in \Sigma \implies {E_1}^c, {E_2}^c \in \Sigma \implies E_1 \cap E_2=({E_1}^c \cup {E_2}^c)^c \in \Sigma$. Moreover, since it is a complete measure space, $E_1\setminus E_2, E_2 \setminus E_1 \in \Sigma$. Having said that, we can write $E_1=(E_1 \setminus E_2) \cup (E_1 \cap E_2)$, the same goes for $E_2$. As $\mu$ is a measure, we have that $\mu(E_1)=\mu(E_1 \setminus E_2) + \mu(E_1 \cap E_2)$, and $\mu(E_2)=\mu(E_2 \setminus E_1) + \mu(E_1 \cap E_2)$.

Again, since $\mu$ is a measure, $0\leq \mu(E_1 \setminus E_2),\mu(E_2 \setminus E_1)\leq \mu(E_1ΔE_2)=0 \implies \mu(E_1 \setminus E_2)=0=\mu(E_2 \setminus E_1)$.

But then, $\mu(E_1)=0+\mu(E_1 \cap E_2)=\mu(E_2)$, which was what we wanted to prove.I would appreciate some help for part a)

$Z_1$ and $Z_1 \setminus Z_2$ are in $\Sigma$. $(Z_1 \cap Z_2) = Z_1 \setminus (Z_1\setminus Z_2)$. The measureable sets form a sigma algebra, right?

 

[mahler1]

$Z_1$ and $Z_1 \setminus Z_2$ are in $\Sigma$. $(Z_1 \cap Z_2) = Z_1 \setminus (Z_1\setminus Z_2)$. The measureable sets form a sigma algebra, right?

Oh, I see, as you've said $(Z_1 \cap Z_2) = Z_1 \setminus (Z_1\setminus Z_2)$, and, $Z_1 \setminus (Z_1\setminus Z_2)=Z_1 \cap (Z_1 \setminus Z_2)^c$, since a $σ-$ algebra is closed under countable intersection, one can deduce from here that $Z_1 \cap Z_2 \in \Sigma$ (for example, I can set $\bigcap_{i \in \mathbb N} E_i$ with $E_1=Z_1, E_2=(Z_1 \setminus Z_2)^c$ and $E_i=A$ for $i\geq 2$). In a similar way, using the fact that a $σ-$algebra is closed under countable union, it is proved that $Z_2=(Z_2 \setminus Z_1) \cup (Z_1 \cap Z_2) \in \Sigma$

Thanks!

 

[Dick]

Oh, I see, as you've said $(Z_1 \cap Z_2) = Z_1 \setminus (Z_1\setminus Z_2)$, and, $Z_1 \setminus (Z_1\setminus Z_2)=Z_1 \cap (Z_1 \setminus Z_2)^c$, since a $σ-$ algebra is closed under countable intersection, one can deduce from here that $Z_1 \cap Z_2 \in \Sigma$ (for example, I can set $\bigcap_{i \in \mathbb N} E_i$ with $E_1=Z_1, E_2=(Z_1 \setminus Z_2)^c$ and $E_i=A$ for $i\geq 2$). In a similar way, using the fact that a $σ-$algebra is closed under countable union, it is proved that $Z_2=(Z_2 \setminus Z_1) \cup (Z_1 \cap Z_2) \in \Sigma$

Thanks!

You're welcome! And you can just say that if C and D are in the sigma algebra then $C \cap D$ is in the sigma algebra. You don't have to spell out the whole countable intersection. Two is countable.