Subsequences of bounded monotonically increasing function

[RBG]

Assume that $\{f_n\}$ is a sequence of monotonically increasing functions on $\mathbb{R}$ with $0\leq f_n(x) \leq 1 \forall x, n$. Show that there is a subsequence $n_k$ and a function $f(x) = \underset{k\to\infty}{\lim}f_{n_k}(x)$ for every $x\in \mathbb{R}$.

(1) Show that some subsequence nk converges for all rational r. This isn't too hard.
(2) Define $f(x) = \underset{r<x}{\sup}\{f(r)\}$ where $r$ is rational and $x$ is arbitrary.
(3) Show $f_{n_k} (x) \to f(x) \forall x$ where $f$ is continuous.

This proof seems sketchy:

If $f$ is continuous then we need to show $|f(x)-f_n(x)|<\epsilon$ for any $x,\epsilon>0$ and sufficiently large $n$. Well, $|f(x)-f_n(x)|=|\underset{r<x}{\sup}\{f(r)\}-\underset{r<x}{\sup} \{f_n(r)\}|=|\underset{r<x}{\sup} \{f(r)-f_n(r)\}|=|\epsilon_2|$ for sufficiently large $n$ by part a). Then changing the form of the epsilon's this would prove the claim.

Is this the right idea?

(4) Show that f has at most countably many discontinuities and so there. is a further subsequence nl along which fnl (x) → f(x) for every x ∈ R.

I showed that $f$ has at most countably many discontinuities, but how does this imply there is a subsequence that converges to $f(x)$ for all $x$? Do I just "throw out" any function that "creates" a discontinuity? By this I mean, if $f$ is discontinuous at $y$ then remove all the functions in the sequence that either have values larger than $f(y+)$ at $y$ or functions with values smaller than $f(y-)$ at $y$.

 

[Samy_A]

Never mind.

 

[Samy_A]

Assume that $\{f_n\}$ is a sequence of monotonically increasing functions on $\mathbb{R}$ with $0\leq f_n(x) \leq 1 \forall x, n$. Show that there is a subsequence $n_k$ and a function $f(x) = \underset{k\to\infty}{\lim}f_{n_k}(x)$ for every $x\in \mathbb{R}$.

(1) Show that some subsequence nk converges for all rational r. This isn't too hard.
(2) Define $f(x) = \underset{r<x}{\sup}\{f(r)\}$ where $r$ is rational and $x$ is arbitrary.
(3) Show $f_{n_k} (x) \to f(x) \forall x$ where $f$ is continuous.

This proof seems sketchy:

If $f$ is continuous then we need to show $|f(x)-f_n(x)|<\epsilon$ for any $x,\epsilon>0$ and sufficiently large $n$. Well, $|f(x)-f_n(x)|=|\underset{r<x}{\sup}\{f(r)\}-\underset{r<x}{\sup} \{f_n(r)\}|=|\underset{r<x}{\sup} \{f(r)-f_n(r)\}|=|\epsilon_2|$ for sufficiently large $n$ by part a). Then changing the form of the epsilon's this would prove the claim.

Is this the right idea?

Can you explain this step:
$|f(x)-f_n(x)|=|\underset{r<x}{\sup}\{f(r)\}-\underset{r<x}{\sup} \{f_n(r)\}|$?
It looks as if you assume here that $f_n$ is continuous in $x$, but that is not necessarily the case.
Also, you don't have to show that $|f(x)-f_n(x)|<\epsilon$ for any $x,\epsilon>0$ and sufficiently large $n$. You have to prove that for the subsequence defined in 1): $|f(x)-f_{n_k}(x)|<\epsilon$ ...
Use that $f$ is monotonically increasing, the continuity of $f$ at $x$, and the denseness of the rational numbers.

(4) Show that f has at most countably many discontinuities and so there. is a further subsequence nl along which fnl (x) → f(x) for every x ∈ R.

I showed that $f$ has at most countably many discontinuities, but how does this imply there is a subsequence that converges to $f(x)$ for all $x$? Do I just "throw out" any function that "creates" a discontinuity? By this I mean, if $f$ is discontinuous at $y$ then remove all the functions in the sequence that either have values larger than $f(y+)$ at $y$ or functions with values smaller than $f(y-)$ at $y$.

You can prove 4) in essentially the same way you constructed the subsequence in 1). That $f$ has at most countably many discontinuities is indeed the key.

 

[RBG]

Thank you!