- #1
RBG
- 14
- 0
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##.
(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##.