The monotone convergence theorem states that any bounded, non-decreasing (or non-increasing) sequence of real numbers converges to its supremum (or infimum). Can we extend this result to functions? I think we could, but I am not sure... I haven't found this anywhere so I'm worried that my idea is wrong. Hopefully someone can guide me:
Suppose $f:\mathbb R \to \mathbb R$ is a bounded and non-decreasing function such that its limit at infinity exists (I am not sure whether the previous conditions suffice for the limit to exist). If we take any non-decreasing sequence $a_n$, then $f(a_n)$ is also a non-decreasing sequence of real numbers which is bounded, so that the monotone convergence theorem guarantees that $f(a_n)$ must converge to its supremum $S$. But, precisely, $\sup f = S$, and since all subsequences of $f$ must converge to the same limit as $f$, then we get $\lim f = \sup f$.
Is this correct?
Yes, you can extend the result for limits of functions at infinity, because
So if $f$ is monotone bounded then every sequence $f(x_n)$ will be monotone bounded thus convergent, then it must converge to $L = \sup f$, and we have $\lim_{x\to+\infty} f(x) = L$.
Edit: Actually it's more instructive to prove directly that $\lim_{x\to+\infty} f(x) = L$: Suppose $f$ is non-decreasing. Given $\epsilon>0$, $L-\epsilon$ is not an upper bound of $\text{Im} f$, so there is a $x_0$ such that $L-\epsilon < f(x_0) \le L$. If $x > x_0$, because $f$ is non-decreasing we have $L-\epsilon < f(x_0) \le f(x) \le L$ then $L-\epsilon < f(x) < L+\epsilon$ so we showed that there is a $x_0$ such that $x>x_0$ implies $|f(x)-L|<\epsilon$, i.e., that $\lim_{x\to+\infty} f(x) = L$.