Theorem 11.33b)
Suppose $f$ is bounded on $[a,b]$. Then $f \in \mathfrak{R}$ (Riemann Integrable) if and
only if $f$ is continuous almost everywhere on $[a,b]$.
The structure of the proof proceeds as follows:
Construct a series of partitions $P_k$, each being a refinement of the previous partition and with adjacent points less than $\frac{1}{k}$ apart. Then if $x \notin \cup_k P_k$, then $L(x) = U(x)$ iff $f$ is continuous at $x$. ( $L_k(x)$ is the (lower) function that corresponds to the infinum of $f(x)$ in each of its intervals partitioned by $P_k$, and $L(x)$ is the limit of $L_k(x)$ as $k \to \infty$. For the upper function $U(x)$ the supremum is used)
Rudin then retrieves a result in the previous part of the proof, namely that $f \in \mathfrak{R}$ iff $L(x) = U(x)$ almost everywhere.
Here is the part that I do not understand – Rudin concludes by saying:
Since the union of the sets $P_k$ is countable, its measure is $0$, and we conclude that $f$ is continuous almost everywhere on $[a,b]$ iff $L(x) = U(x)$ almost everywhere, hence iff $f \in \mathfrak{R}$.
I understand the two premises separately, but I don't see how they combine to give the conclusion of "iff $f \in \mathfrak{R}$". Can someone explain this?
Best Answer
It helps to write out in set language. The assertion "if $x\not\in\cup_kP_k$, then $L(x)=U(x)$ iff $f$ is continuous at $x$" translates to: $$ N^c\cap\{x: L(x)=U(x)\} = N^c\cap \{x: \text{$f$ is continuous at $x$}\}\tag1 $$ where for brevity we write $N:=\cup_kP_k$, a set of measure zero. Equivalently: $$ N\cup\{x: L(x)\ne U(x)\} = N\cup \{x: \text{$f$ is discontinuous at $x$}\}\tag2 $$
The result Rudin retrieves is:
Since $N$ has measure zero, the RHS is equivalent to:
and by (2) is equivalent to
and, again since $N$ has measure zero, is equivalent to: