# measurable functions

• Sep 28th 2009, 02:05 AM
robeuler
measurable functions
Let f and g be measurable functions from X to the extended reals. Prove that the sets

A={x:f(x)<g(x)}
and
B={x:f(x)=g(x)}
are measurable.

And then prove that the set of points at which a sequence of measurable real valued functions converges is measurable.

To show that A is measurable I am trying to work with intersections. I know that f is measurable iff {(a,inf]} is measurable for all a in R. Similarly for g. Can I simply take the intersection of such sets to get A? Is the intersection of measurable sets measurable?

Can I do something similar for B?

Assuming A and B are measurable, I suppose I have to use that to show the second part of the question, but I dont see how.
• Sep 28th 2009, 05:00 AM
putnam120
1) Given the sigma algebra (the collection of measurable sets), $\displaystyle A$, then if $\displaystyle X,Y\in A$ you have that $\displaystyle X^c,Y^c\in A$. Next $\displaystyle X\cap Y=\left(X^c\cup Y^c\right)^c$. So $\displaystyle X\cap Y\in A$.

2) Let $\displaystyle E_(n,m,k)=\left\{x:|f_n(x)-f_m(x)|<\frac{1}{k}\right\}$. Then $\displaystyle \bigcap_{k=1}^\infty\bigcup_{N=1}^\infty\bigcap_{n =N}^\infty\bigcap_{m=N}^\infty E(n,m,k)$ is measurable. Now just show it is the same as the set of points where $\displaystyle f_n$ converges. Oh and you need to justify that $\displaystyle E(n,m,k)$, and the final set (with the intersections and unions) are both measurable.
• Sep 28th 2009, 08:50 AM
problem
For set A,try this A=$\displaystyle \bigcup_{r\in Q}(\{x:f(x)<r\}\bigcap \{x:g(x)>r\})$,Q is the set of rational numbers.
set B can be expressed as $\displaystyle \{x:f(x)\le g(x)\} \bigcap \{x:g(x)\le f(x)\}$
Quote:

And then prove that the set of points at which a sequence of measurable real valued functions converges is measurable.
The required set is $\displaystyle \{x:lim inf f_n(x)=lim sup f_n(x)\}$ where it follows that $\displaystyle lim inf f_n$ and $\displaystyle lim sup f_n$are measurable functions.