Prove This Set is Measurable

• December 19th 2010, 07:19 PM
Guy
Prove This Set is Measurable
Hey all,

Problem: Let $\{f_n\}$ be a sequence of measurable functions. Prove that the set of points at which $\{f_n\}$ converges is a measurable set.

My Attempted Solution: We know that $\limsup f_n$ and $\liminf f_n$ are measurable functions; hence $g = \limsup f_n - \liminf f_n$ is measurable. The set of values at which $\{f_n\}$ converges is equal to $\{x: g(x) = 0\}$ which is measurable because $g$ is measurable. Q.E.D.

Does this work? I tried to check my answer online, but the methods I saw differ pretty significantly from this, which makes me question whether or not this works.
• December 20th 2010, 12:24 AM
Opalg
Quote:

Originally Posted by Guy
Hey all,

Problem: Let $\{f_n\}$ be a sequence of measurable functions. Prove that the set of points at which $\{f_n\}$ converges is a measurable set.

My Attempted Solution: We know that $\limsup f_n$ and $\liminf f_n$ are measurable functions; hence $g = \limsup f_n - \liminf f_n$ is measurable. The set of values at which $\{f_n\}$ converges is equal to $\{x: g(x) = 0\}$ which is measurable because $g$ is measurable. Q.E.D.

Does this work? I tried to check my answer online, but the methods I saw differ pretty significantly from this, which makes me question whether or not this works.

That is a good proof. Of course, it assumes the result about limsup and liminf being measurable, which is where the hard work is buried.

Edit. On second thoughts, you need to do a bit more work to deal with points where limsup and liminf are both $+\infty$ or both $-\infty$. At those points, g(x) is not defined.
• December 20th 2010, 07:23 AM
Guy
Dang, you're right. How about this as a quick fix:

Restrict $g$ to be defined only where $\limsup f_n \ne -\infty$ and $\liminf f_n \ne \infty$. Then $\{x: g(x) = 0\}$ is still measurable (I think this is straightforward from the fact that liminf and limsup are still measurable when restricted to the set where this guy is defined), and the set of interest is given by

$\{x:g(x) = 0\} \cup (\limsup f_n)^{-1} (\{-\infty\}) \cup (\liminf f_n)^{-1} (\{\infty\})$

which is a union of measurable sets. Does that fix it up?
• December 20th 2010, 08:11 AM
Opalg
Quote:

Originally Posted by Guy
Dang, you're right. How about this as a quick fix:

Restrict $g$ to be defined only where $\limsup f_n \ne -\infty$ and $\liminf f_n \ne \infty$. Then $\{x: g(x) = 0\}$ is still measurable (I think this is straightforward from the fact that liminf and limsup are still measurable when restricted to the set where this guy is defined), and the set of interest is given by

$\{x:g(x) = 0\} \cup (\limsup f_n)^{-1} (\{-\infty\}) \cup (\liminf f_n)^{-1} (\{\infty\})$

which is a union of measurable sets. Does that fix it up?

That argument looks fine, but I would give the "set of interest" as $\{x:g(x) = 0\}$ (without the other bits). You normally only say that a sequence converges if it has a finite limit.