I am really stuck on the following, I don't know how to start. Can somebody please help?
Suppose f:[a,b]->R is a weakly increasing function, i.e. u<v gives f(u)<=f(v). Show that f is the limit of a sequence of step functions.
I can imagine that the sequence of step functions f_n would converge to f. However I do not know how to prove it. Can you please help? Here is where I am stuck.
I pick e>0, I now need to show that there exists N so that for all n>N,
|f_(x)-f(x)|<e. However, as I don't know which specific function f is I am really struggling. How do I pick n?
You may find it a bit easier to divide the range of the function (the interval [f(a),f(b)]) into subintervals of length less than . If J is one of those subintervals, use the weakly increasing condition to show that the set is an interval. Let y_J be the value that f takes at some point of J. Then the function that takes the constant value on J differs from f by an amount less than throughout J. Doing this for each of the subintervals of [f(a),f(b)], you get a step function that is always close to f.
I think I have got it. Is this right?
pick n so that (f(b)-f(a))/n <e.
let p_i= f(a) + i(f(b)-f(a))/n (i between 1 and n-1)
let J=(p_i, p_i+1)
pick any z,y in [a,b] such that f(z)=p_i, f(y)=p_i+1
let g(x)=0.5*(p_i+1 +p_i)/2 for all x in [z,y].
for any x in [p_i,p_i+1], |g(x)-f(x)|<=|p_i+1-f(x)|<=|p_i+1-p_i|<e