- #1
mahler1
- 222
- 0
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)
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)