99% of math majors cannot prove this

Proposition 3, part (b). From MIT's 2005 stochastic processes course. A few different ways to do it, most elegant wins.

You already tried asking this in the other thread, fuck off.

Its complement is the countable union of the Borel sets B(0,n,n), where n runs over the natural numbers.

And try asking this kind of stuff on math.stackexchange.com in stead, it has actual mathematicians and far fewer stoned wankers.

Oh my goodness, this is so easy. Seriously, it is extremely trivial.

I won't solve it here or anywhere else because I don't want the mathfags to get anymore depressed, but trust me, I know how to do it.

I'll use m and n separately just for clarity:

So if we take the countable union over all n in N (naturals) of B(0,n,m) we get the set of all functions between -n and n in an interval [0,m) for some n. Then taking the intersection of all m in N, we get the set of all functions between -n and n for some n. So I guess this is really the set of all bounded functions.
So, is the set of all bounded functions equivalent to the set of functions with limit < infinity?

fuck off, i like having actual questions on Veeky Forums

>So, is the set of all bounded functions equivalent to the set of functions with limit < infinity?

the question is talking about continuous functions, so yeah, i think you got it.

Thanks lads, got it out in the end I think.

wait, what about something like x(t) = t*sin(t).

this isn't bounded

yeah he did it in the wrong order. for fixed n, take the intersection over all m of B(0,n,m) (aka "all functions bounded by "n") giving B(0,n). now take the union over all n of B(0,n), aka "all bounded functions"

what i'm saying is you can't just take the complement of all bounded functions, because there are unbounded functions whose limit is not infinity

So what about if you instead show that the set of all unbounded continuous functions are in B. Are all unbounded continuous functions limits at infinity = infinity (or -infinity)?

>Are all unbounded continuous functions limits at infinity = infinity (or -infinity)?

correct me if i'm wrong, but x(t) = t*(sin(t)+1) for instance has no limit at infinity. it just continually goes up and down.

Proof: Think

This is from early in the course, so the answers are restatements of the prerequisites.

(b) field definition

(c) Borel measure definition

>he didn't go for the fermat meme to get the same point across but not come on as much of an arrogant nigger
I have discovered a truly marvelous proof of how much of a fag you are, but the character limit is too small to contain it

dumb question does [0,infty) indicate the domain?

Yes
Define [eqn]G(n, T) = \{ f \in C[0, \infty) \mid f(t) \geq n \ \forall \ t \geq T\}.[/eqn]
A proof similar to that of 3(a) shows that [math]G(n, T) \in \mathcal{B}[/math] for all [math]n, T \in \mathbb{Z}[/math].
Define [eqn]H(n) = \bigcup_{T \in \mathbb{Z}^+} G(n, T).[/eqn]
The set of all functions in [math]C[0, \infty)[/math] with limit [math]\infty[/math] is [eqn]\bigcup_{n \in \mathbb{Z}^+} H(n).[/eqn]

wouldn't that union also include bounded functions?

Woops. That last union should be an intersection.