# Rudin's PMA Theorem 11.33b

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?

• I'm confused. Four lines above, you say that Rudin has retrieved the fact that $f\in\mathscr R$ iff $L(x)=U(x)$ a.e. Commented Apr 25, 2020 at 19:24
• @TedShifrin What's a.e? Commented Apr 25, 2020 at 19:28
• Sorry. almost everywhere. Commented Apr 25, 2020 at 19:37
• @TedShifrin yup Rudin did, but that doesn't necessarily give the conclusion right? Commented Apr 25, 2020 at 19:41
• Can you state very specifically what your question is? Commented Apr 25, 2020 at 20:56

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:

$$f\in{\mathfrak R}$$ if and only if $$\{x: L(x)\ne U(x)\}$$ has measure zero

Since $$N$$ has measure zero, the RHS is equivalent to:

$$N \cup \{x: L(x)\ne U(x)\}$$ has measure zero

and by (2) is equivalent to

$$N \cup \{x: \text{f is discontinuous at x}\}$$ has measure zero

and, again since $$N$$ has measure zero, is equivalent to:

$$\{x: \text{f is discontinuous at x}\}$$ has measure zero.

• Thank you for this clear explanation! Commented Apr 26, 2020 at 5:11

For necessity, Rudin's proof is based on results related to Lebesgue integration. In particular, the result that if $$\int g = \int h$$ and $$g\leq h$$, then $$g=h$$ a.s. Here is more or less Rudin's argument:

Choose partitions $$\mathcal{P}_n\subset\mathcal{P}_{n+1}$$ such that $$U(f,\mathcal{P}_n)-L(f,\mathcal{P}_n)<1/n$$. For each partition $$\mathcal{P}_n$$, let $$m_{n,k}=\inf\{f(t):t\in[t_{n,k-1},t_{n,k}]\}$$ and $$M_{n,k}=\sup\{f(t):t\in[t_{n,k-1},t_{n,k}]\}$$. Let $$g_n$$ and $$h_n$$ be defined by $$g_n(a)=h_n(a)$$; and $$g_n(t)=m_{n,k}$$, $$h_n(t)=M_{n,k}$$ on $$t\in(t_{n,k-1},t_{n,k}]$$. Clearly, $$g_n\leq g_{n+1}\leq f\leq h_{n+1}\leq h_n$$ on $$[a,b]\setminus\mathcal{P}_n$$, and $$\int_{[a,b]}g_n=L(f,\mathcal{P}_n)\leq U(f,\mathcal{P}_n)=\int_{[a,b]}h_n$$.

Dominated convergence implies $$\int_{[a,b]}g(x)dx=\int_{[a,b]}h(x)dx=A(f)$$; Thus, since $$g=\lim_ng_n\leq f\leq \lim_nh_n=h$$, then $$g=f=h$$ a.s. Let $$\mathcal{D}=\{t\in[a,b]:g(t). Then, $$f$$ is continuous at every point $$x\notin\bigcup_n\mathcal{P}_n\cup \mathcal{D}$$.

For sufficiency, I only sketch the proof. Decompose the set of discontinuities $$D$$ as the union of sets where the modulus of discontinuity is larger that say $$\frac1n$$, i.e. $$D=\bigcup_n\{x\in [a,b]:\omega_f(x)\geq\frac1n\}$$ where $$\omega_f(x)$$ is the modulus of continuity of $$f$$ at $$x$$. If $$\lambda(D)=0$$ then all of $$\lambda\{x:\omega_f(x)\geq\frac1n\}=0$$ Then one may argue that for any $$n$$, there is $$\delta_n>0$$ such that whenever $$I\subset[a,b]$$ is a subinterval with $$\lambda(I)<\delta$$ and $$I\cap D=\emptyset$$, then $$\sup\{|f(x)-f(y)|:x,y\in I\}\leq\frac1n$$. Then use partitions that the subintervals that cover the set $$D_n=\{x:\omega_f(x)\geq\frac1n\}$$ have lengths that add up to something less than $$\frac1n$$ (this is possible since $$\lambda(D_n)=0$$) and the each of the remaining subintervals have length less than $$\delta_n$$. Then $$U(f,P)-L(f,P)< (M-m+b-a)/n$$, where $$M=\sup f$$ and $$m=\inf f$$.

Not to change one book for another, but a full argument, using only the notion of a set of measure zero, can be found in the book of Apostol's Mathematical analysis, section 7.26.