# Would someone clarify this for me please? (from Rudin's analysis)

## Main Question or Discussion Point

Theorem: Suppose f is bounded on [a,b], f has only finitely many points of discontinuity on [a,b], and α is continuous at every point at which f is discontinuous. Then f is in R(α).

Proof: Let ε>0 be given. Put M=sup|f(x)|, let E be the set of points at which f is discontinuous. Since E is finite and α is continuous at every point of E, we can cover E by finitely many disjoint intervals [uj, vj] in [a,b] such that the sum of the corresponding differences α(vj)-α(uj) is less than ε.

Why this is true? I understand that we can cover E by those finitely many disjoint intervals, but I don't understand why we could cover E in such a way the sum of the corresponding differences α(vj)-α(uj) would be less than epsilon.

Any help would be appreciated.

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Theorem: Suppose f is bounded on [a,b], f has only finitely many points of discontinuity on [a,b], and α is continuous at every point at which f is discontinuous. Then f is in R(α).

Proof: Let ε>0 be given. Put M=sup|f(x)|, let E be the set of points at which f is discontinuous. Since E is finite and α is continuous at every point of E, we can cover E by finitely many disjoint intervals [uj, vj] in [a,b] such that the sum of the corresponding differences α(vj)-α(uj) is less than ε.

Why this is true? I understand that we can cover E by those finitely many disjoint intervals, but I don't understand why we could cover E in such a way the sum of the corresponding differences α(vj)-α(uj) would be less than epsilon.

Any help would be appreciated.

You can always choose a finite number of intervals (or even an infinite countable number of them) s.t. that their

lenghts' sum is less than any predefined positive quantity...

For example, the sum of lengths of $\,\,[a_n,b_n]\,\,,\,\,n=1,2,...k\,\,$ is less that $\,\,\epsilon>0\,\,$ if $\,\,b_n-a_n\leq\frac{\epsilon}{2^n}\,\,$.

BTW, what is that "a" and what is R(a)??

DonAntonio

You can always choose a finite number of intervals (or even an infinite countable number of them) s.t. that their

lenghts' sum is less than any predefined positive quantity...

For example, the sum of lengths of $\,\,[a_n,b_n]\,\,,\,\,n=1,2,...k\,\,$ is less that $\,\,\epsilon>0\,\,$ if $\,\,b_n-a_n\leq\frac{\epsilon}{2^n}\,\,$.

BTW, what is that "a" and what is R(a)??

DonAntonio
Yes, you can always choose a finite number of intervals that their lenghts' sum is less than any given positive quantity, but that's not the case here.
Here we are choosing a finite number of intervals, that when we apply α (an increasing function), the sum of the corresponding differences α(vj)-α(uj) is less than any given epsilon. So, it's not like the case you said I think.

R(α) is the set of all Riemann-stieljes integrable functions. All functions like ∫fdα that are integrable. (Rudin assumes α is an increasing function).

Yes, you can always choose a finite number of intervals that their lenghts' sum is less than any given positive quantity, but that's not the case here.
Here we are choosing a finite number of intervals, that when we apply α (an increasing function), the sum of the corresponding differences α(vj)-α(uj) is less than any given epsilon. So, it's not like the case you said I think.

R(α) is the set of all Riemann-stieljes integrable functions. All functions like ∫fdα that are integrable. (Rudin assumes α is an increasing function).

Oh, I see...but then it is also clear: at every discontinuity point $\,\,x_0\,\,$ of f, where g (this is your function a) is continuous, we have that $$\,\,\exists \delta>0\,\,s.t.\,\,|x-x_0|<\delta\Longrightarrow |g(x)-g(x_0)|<\epsilon$$ and now you can choose your intervals within $\,\,|x-x_0|<\delta_m\,\,,\,\,\delta_m:=\,\,$ the minimal such number among all that finite ammount of them.

DonAntonio

Now it makes sense. Thanks.