# Thread: Little question - measure theory

1. ## Little question - measure theory

I had this little thought bothering me:

Let $\displaystyle (X,\mathcal A,m)$ be a finite measure space , two functions $\displaystyle f,g$, and $\displaystyle \mathcal B$ a sub $\displaystyle \sigma$- algebra.

If $\displaystyle f=g$ a.e. and $\displaystyle g$ is $\displaystyle \mathcal B$-measurable,

is $\displaystyle f$ $\displaystyle \mathcal B$-measurable too?

Im thinking yes but im not sure how to prove it.

Any ideas?

thnx

2. Originally Posted by mabruka
I had this little thought bothering me:

Let $\displaystyle (X,\mathcal A,m)$ be a finite measure space , two functions $\displaystyle f,g$, and $\displaystyle \mathcal B$ a sub $\displaystyle \sigma$- algebra.

If $\displaystyle f=g$ a.e. and $\displaystyle g$ is $\displaystyle \mathcal B$-measurable,

is $\displaystyle f$ $\displaystyle \mathcal B$-measurable too?

Im thinking yes but im not sure how to prove it.

Any ideas?

thnx
Yes it is

Let O be an open set

$\displaystyle g^{-1}(O) =$ {$\displaystyle x$ ; $\displaystyle g(x) \in O$} = [{$\displaystyle x$ ; $\displaystyle f(x) \in O$} \ {$\displaystyle x$ ; $\displaystyle f(x) \neq g(x)$}] $\displaystyle \cup$ {$\displaystyle x$ ; $\displaystyle g(x) \in O$ & $\displaystyle f(x) \neq g(x)$}

But everyone of these sets is measurable (see why?), so then $\displaystyle g^{-1}(O)$ is too. Thus g is measurable.

3. You reversed the roles of f and g but it is ok, lets do it your way .

i was considering that decomposition of the inverse mapping but got stuck in the following part.

Lets call $\displaystyle E$ the exceptional set where f and g differ , $\displaystyle m(E)=0$ and $\displaystyle E=\{x: f(x)\not =g(x)\}$

First question:

The left side of $\displaystyle f^{-1}(O)$ is

$\displaystyle f^{-1}(O)\cap E^c =g^{-1}(O)\cap E$ which is obviously in $\displaystyle \mathcal A$ but i am not sure why it must be in $\displaystyle \mathcal B$ since $\displaystyle \mathcal B \subset \mathcal A$

The right side

$\displaystyle f^{-1}(O)\cap E \subset E$ ... and thus
$\displaystyle m(f^{-1}(O)\cap E \subset E)=0$ but i dont know what to do next..

thank you southprkfan1

4. Originally Posted by mabruka
You reversed the roles of f and g but it is ok, lets do it your way .

i was considering that decomposition of the inverse mapping but got stuck in the following part.

Lets call $\displaystyle E$ the exceptional set where f and g differ , $\displaystyle m(E)=0$ and $\displaystyle E=\{x: f(x)\not =g(x)\}$

First question:

The left side of $\displaystyle f^{-1}(O)$ is

$\displaystyle f^{-1}(O)\cap E^c =g^{-1}(O)\cap E$ which is obviously in $\displaystyle \mathcal A$ but i am not sure why it must be in $\displaystyle \mathcal B$ since $\displaystyle \mathcal B \subset \mathcal A$

The right side

$\displaystyle f^{-1}(O)\cap E \subset E$ ... and thus
$\displaystyle m(f^{-1}(O)\cap E \subset E)=0$ but i dont know what to do next..

thank you southprkfan1
My Bad. I made 2 mistakes. That one, and more importantly, I didn't notice you had 2 different - algebras.

... Let me get back to you, sorry

5. Well it turns out that even when considering only one sigma algebra like you did the statement may not be true if the measure space $\displaystyle (X,\mathcal F, m)$ is not complete in the sense that

$\displaystyle (X,\mathcal F, m)$ is complete if the all the null sets are measurable.
$\displaystyle \mathcal N = \{A\subset X: \exists B\in \mathcal F \text{ such that } A\subset B, m(B)=0 \}\subset \mathcal F$

So well, lets add the "$\displaystyle (X,\mathcal F, m)$ complete" to the hypothesis.

The case when we have only one sigma algebra is true.
The other one remains yet unclear.

6. Originally Posted by mabruka
Well it turns out that even when considering only one sigma algebra like you did the statement may not be true if the measure space $\displaystyle (X,\mathcal F, m)$ is not complete in the sense that

$\displaystyle (X,\mathcal F, m)$ is complete if the all the null sets are measurable.
$\displaystyle \mathcal N = \{A\subset X: \exists B\in \mathcal F \text{ such that } A\subset B, m(B)=0 \}\subset \mathcal F$

So well, lets add the "$\displaystyle (X,\mathcal F, m)$ complete" to the hypothesis.

The case when we have only one sigma algebra is true.
The other one remains yet unclear.

I'll be honest, I'm completely stuck .

I had never of a measure space being complete. I was taught that all zero sets are measurable, period. Very interesting.

7. Well the completeness is just a technical aspect that most of the time is overlooked because if you have a measure $\displaystyle m$ over $\displaystyle \mathcal F_1$ which is not complete you can always extend it to a measure $\displaystyle M$ over a bigger sigma algebra $\displaystyle \mathcal F_2$

satisfying that $\displaystyle M|_{\mathcal F_1} =m$

This is done using Carathedory's Extension theorem. In fact if the initial measure is finite then the extension is unique.