Differentiability of Monotone Function's: Lebesgue's Theorem

[Artusartos]

a)Does convergence imply being properly defined? So would it not be properly defined if it was divergent?

b)I am having trouble why the last part (in the attachment) says, "Then, by (1), [itex]f(x_0) - f(x) \geq dfrac{2^k} for all $x < x_0$." But does (1) tell us that it's "equal" instead of "greater than or equal to"?<br /> <br /> Thanks in advance[/itex]

 

[micromass]

a) We define the function

$$f(x) = \sum_{q_n\leq x} \frac{1}{2^n}$$

But if the series in the RHS diverges for a certain $x$. Then we would have $f(x)=+\infty$. But $+\infty$ is not a real number. We want $f(x)$ to be a real number for any $x$. This is only satisfied if the series converge.

So yes, if we say that $f(x)$ is well-defined, then that actually means that the series converge.

b) I'm not quite understanding why you expect the equality to hold. You have

$$f(q_k) - f(x) = \sum_{x<q_n\leq q_k} \frac{1}{2^n} \geq \frac{1}{2^k}$$

The inequality in the end of course holds because $1/2^k$ is a term in the sum on the LHS. So the inequality means that you drop every term in the sum except $1/2^k$. I don't get why you think that this should be an equality.

 

[Artusartos]

a) We define the function

$$f(x) = \sum_{q_n\leq x} \frac{1}{2^n}$$

But if the series in the RHS diverges for a certain $x$. Then we would have $f(x)=+\infty$. But $+\infty$ is not a real number. We want $f(x)$ to be a real number for any $x$. This is only satisfied if the series converge.

So yes, if we say that $f(x)$ is well-defined, then that actually means that the series converge.

b) I'm not quite understanding why you expect the equality to hold. You have

$$f(q_k) - f(x) = \sum_{x<q_n\leq q_k} \frac{1}{2^n} \geq \frac{1}{2^k}$$

The inequality in the end of course holds because $1/2^k$ is a term in the sum on the LHS. So the inequality means that you drop every term in the sum except $1/2^k$. I don't get why you think that this should be an equality.

Thanks, but I have one more question...

$$f(q_k) - f(x) = \sum_{x<q_n\leq q_k} \frac{1}{2^n} \geq \frac{1}{2^n}$$, right? But then how do we have $|f(x)-f(x_0)| = \frac{1}{2^n} < \frac{1}{2^n}$?

I don't know, I think I'm confused about what n represents.

 

[Bacle2]

Basically, for any finite n, you can construct a ball about xo that excludes all the

elements Sn:={q1,...,qn}, since min{ |xo-qi| i=1,2,..n} :=d >0

meaning, for finitely-many elements qi , the distance from xo to any of them is

non-zero, and the since the set Sn is finite, the set of distances from xo to any member in Sn has a minimum ( no need to worry about infs.) ,

say that minimum is called d.

Then, within the ball, say, B(x,d/2) , the only possible elements of C in the ball

are {q_(n+1),q_(n+2),...} . Now, by definition of f(x), can

you see why the difference can be made as small as possible (think of the elements of C between x and xo in the ball)?

 

[Artusartos]

Basically, for any finite n, you can construct a ball about xo that excludes all the

elements Sn:={q1,...,qn}, since min{ |xo-qi| i=1,2,..n} :=d >0

meaning, for finitely-many elements qi , the distance from xo to any of them is

non-zero, and the since the set Sn is finite, the set of distances from xo to any member in Sn has a minimum ( no need to worry about infs.) ,

say that minimum is called d.

Then, within the ball, say, B(x,d/2) , the only possible elements of C in the ball

are {q_(n+1),q_(n+2),...} . Now, by definition of f(x), can

you see why the difference can be made as small as possible (think of the elements of C between x and xo in the ball)?

Thanks a lot, but I'm not really familiar with "balls"...

 

[Bacle2]

An open ball B(x,r) in the real line , i.e., x is any real number and d is a positive real is the set:

B(x,r):={ y in Reals : |x-y|<d }