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

To prove this, we cover E, the set of points of discontinuities, by finitely many disjoint intervals $[u_j, v_j]\subset [a, b]$. Furthermore we place these intervals in such a way that every point of $E\cap (a, b)$ lies in the interior of some $[u_j, v_j]$. Define $K=[a, b]\setminus \bigcup (u_j, v_j)$. Then it is said that K is compact and $f$ is continuous on K. How?

If $f$ is discontinuous at $a$ then $a=u_1$ and $a\in K$ so that $f$ is not continuous on $K$. So in this case, I think, we have to remove the interval $[u_1, v_1)$ from $[a, b]$ instead of $(u_1, v_1)$. Is this correct?

matthew
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    Could you state the theorem and tell us what $K, u_1, v_1$ are so those who don't have access to baby rudin can understand the question? – Rachmaninoff Mar 28 '14 at 02:16
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    Thanks for the edit. If $f$ is discontinuous at $a$, then $a\in K$ (as you indicate), but $a$ is an isolated point of $K$, because $a$ belongs to some $[u_j,v_j]\subset[a,b]$, so $u_j=a$, and we removed from $K$ the whole interval $(a,v_j)$. Being an isolated point of $K$, the restriction of $f$ to $K$ is continuous at $a$. Note that we only look at this restriction in what remains of the argument. – Andrés E. Caicedo Mar 29 '14 at 01:38

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Turning my comment into an answer:

(Just in case: $K$ is compact, since it is the intersection of the compact set $[a,b]$ with the closed set that is the complement of $\bigcup_j(u_j,v_j)$. And the restriction of $f$ to $K$ is continuous at all points of $K$, by assumption, except perhaps at $a$ or $b$.)

If $f$ is discontinuous at $a$, then $a\in K$ (as you indicate), but $a$ is an isolated point of $K$, because $a$ belongs to some $[u_j,v_j]\subset [a,b]$, so $u_j=a$, and we removed from $K$ the whole interval $(a,v_j)$. Being an isolated point of $K$, the restriction of $f$ to $K$ is continuous at $a$. Note that we only look at this restriction in what remains of the argument.

Andrés E. Caicedo
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