# Understanding Rudin's proof that a Riemann integrable function is measurable

In the book "Principles of Mathematical Analysis" by Walter Rudin, he proves the following theorem (slightly reworded),

Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and $$\int_a^b f \ dx = \mathscr{R} \int_a^b f \ dx$$ Where $\mathscr{R} \int$ denotes the Riemann integral, while $\int$ denotes the Lebesgue integral.

Proof Suupose $f$ is bounded. Then there exists a sequence $\{P_k\}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and $$\lim_{k\rightarrow\infty} L(P_k,f) = \mathscr{R}\underline{\int_a^b} f \ dx, \quad \lim_{k\rightarrow\infty} U(P_k,f) = \mathscr{R}\overline{\int_a^b} f \ dx.$$ Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k=\{a=x_0<x_1<\dots<x_n=b\},$ these are defined as, $$L(P_k,f) = \sum_{i=1}^n (x_i-x_{i-1})m_i, \quad U(P_k,f) = \sum_{i=1}^n (x_i-x_{i-1})M_i,$$ where $M_i = \sup_{x\in[x_{i-1},x_i]} f(x)$ and $m_i = \inf_{x\in[x_{i-1},x_i]} f(x).$

We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x \in(x_{i-1},x_i],$ $1\leq i \leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $x\in [a,b],$ $$L(P_k,f) = \int_a^b L_k \ dx, \quad U(P_k,f) = \int_a^b U_k \ dx,$$ and $$L_1(x) \leq L_2(x) \leq \dots \leq f(x) \leq \dots \leq U_2(x) \leq U_1(x).$$ There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x \in [a,b],$ $$L(x) \leq f(x) \leq U(x),$$ and by the monotone convergence theorem, $$\int_a^b L(x) \ dx = \mathscr{R} \underline{\int_a^b} f \ dx, \quad \int_a^b U(x) \ dx = \mathscr{R} \overline{\int_a^b} f \ dx.$$ Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) \leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].

Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.

I've understood the proof up until the last sentence, but I don't understand how you can conclude that $f$ is measurable (the part I bolded).

I know it is a standard result that working with the Lebesgue measure on $[a,b],$ if $f$ is measurable and $f(x)=g(x)$ almost everywhere on, then $g$ is also measurable (and that this result also holds more generally). However when trying to prove this, I needed to show that a subset of a measure zero set is measurable. While I know this is true for the Lebesgue measure, I'm struggling to prove it using the definition of measurable sets provided by Rudin.

To be brief, Rudin defines a set to be finitely $\mu$-measurable if there exists a sequence of elementary sets which converges to it, in the sense that the outer measure of the symmetric difference converges to $0.$ A set if *$\mu$-measurable if there exists a sequence of finitely $\mu$-measurable sets that converge to it.

I'm also wondering if there's an easier way to prove this assertion, rather than referring to the more general result.

Edit in response to hardmath's response: To prove that $f$ is measurable using the fact that $L$ is and $f=L$ almost everywhere, I initially thought to do the following,

Let $E = \{ x \mid f(x) \neq L(x) \},$ then $E$ is measurable and $m(E) = 0,$ where $m$ is the Lebesgue measure. We wish to show that for every $c \in \Bbb R,$ the set $\{ x \mid f(x)>c\}$ is measurable (this is how Rudin defines measurable functions). So given an arbitrary $c,$ we have, $$f^{-1}((c,\infty]) = \left(f^{-1}((c,\infty]) \cap E \right) \ \cup \ \left(L^{-1}((c,\infty]) \cap E^C \right),$$ since $f$ and $L$ agree on $E.$ The second term is measurable, but I'm not sure how to one can conclude that the first is.

• Note that $f$ is a measurable function, not a measurable set, right? The conclusion is that since $f(x)$ agrees with the measurable functions $L(x),U(x)$ "almost everywhere" on $[a,b]$, then $f(x)$ itself is a measurable function. The difficulty you seem to be having is connecting "almost everywhere" with the notion that the functions all agree except on a set of measure zero. Is that where your difficulty lies? – hardmath Jul 18 '15 at 13:30
• @hardmath: Not quite. I get that the set of such points in measure zero, but to prove that inverse images of $f$ is measurable it seems like you need the subsets of this set to also be measurable. I'll edit my answer to add details. – ktoi Jul 18 '15 at 13:39
• Not every measure has the property that arbitrary subsets of sets of measure zero are again measurable, but as you note the Lebesgue measure has this property. Measures with this property are called complete. More about this in this previous Question. – hardmath Jul 18 '15 at 14:41
• @hardmath: Indeed, the difficulty I was having was to prove that the Lebesgue measure is complete. I think I've found a way of finishing the proof of the theorem though (using a different approach), which I've posted as an answer. – ktoi Jul 18 '15 at 14:47

A slightly different way to see that $f$ is mensurable:
$\{L_k\}$ (and also $\{U_k\}$) is a sequence of simple functions converging almost everywhere to $f$. So $f$ is mensurable.
Let $c \in \Bbb R.$ Since each $L_k$ is measurable, if we let $A_k = \{ x \mid L_k(x) >c\}$ then these sets are measurable. Further since each set is bounded, they are finitely $\mu$-measurable (known result). If we define for any sets $X, Y,$ $$d(X,Y) = m^*( (X\setminus Y) \cup (Y \setminus X))$$ Where $m^*$ is the Lebesgue outer measure. Then if $A = \{ x \mid L(x) > c \}$ and $B = \{ x \mid f(x) > c \},$ $$d(A_k,B) \leq d(A_k,A) + d(A,B) \rightarrow 0,$$ since $d(A,B)=0,$ by monotonicity of $m^*.$ Thus it follows that $B$ is measurable, by the definition provided in the book.