# Thread: Choosing a sigma algebra such that a function is measurable

1. ## Choosing a sigma algebra such that a function is measurable

I am using the following definition of "measurable." Suppose we have a set $\Omega$ with an associated sigma-algebra $F$ of subsets. A function $f: \Omega \rightarrow R$ is measurable w.r.t. $F$ if: for all $c \in R$, the set $\{\omega \in \Omega | f(\omega) < c\} = f^{-1}((-\infty,c))$ belongs to $F$.

So I choose $\Omega = R$ (the reals) and let the function $f$ be the identity function $f(x) = x$. I hope that this innocent bijective function is measurable. And for all $c$, we would have: $f^{-1}((-\infty,c)) = (-\infty,c)$.

But "measurable" is defined in the context of "with respect to" a sigma-algebra, right? So this requires me to choose a sigma-algebra, no?

Would choosing the trivial sigma-algebra $F= \{\Omega,\emptyset\} = \{R,\emptyset\}$ work?

What other possible sigma-algebras $F$ of $\Omega$ could we choose to keep $f$ measurable w.r.t. $F$?

Thanks!

2. Originally Posted by Last_Singularity
Would choosing the trivial sigma-algebra $F= \{\Omega,\emptyset\} = \{R,\emptyset\}$ work?

It does not work because

$f^{-1}(-\infty,c)=}(-\infty,c)\not\in F$

What other possible sigma-algebras $F$ of $\Omega$ could we choose to keep $f$ measurable w.r.t. $F$?

Any sigma-algebra which contains to the sigma-algebra generated by the family:

$S=\{(-\infty,c):c\in\mathbb{R}\}$

Fernando Revilla

3. Doesn't a sigma-algebra need to be closed under complements? As in, if $F$ is a sigma-algebra and $A \in F$, then $A^C = \Omega - A \in F$.

In this case, if I let $c$ be given, $A=(-\infty,c_A)$ creates $A^C = R - (-\infty,c_A) = [c_A,\infty)$, which is not in $F = \{(-\infty,c): c \in R\}$, right?

I must be mis-understanding the definition...

4. Originally Posted by Last_Singularity
Doesn't a sigma-algebra need to be closed under complements? As in, if $F$ is a sigma-algebra and $A \in F$, then $A^C = \Omega - A \in F$.

In this case, if I let $c$ be given, $A=(-\infty,c_A)$ creates $A^C = R - (-\infty,c_A) = [c_A,\infty)$, which is not in $F = \{(-\infty,c): c \in R\}$, right?

I must be mis-understanding the definition...
The $\sigma$-algebra generated by a family $\mathcal{F}$ is defined as the smallest $\sigma$-algebra which contains $\mathcal{F}$.