In previous years, Lebesgue integration was alluded to as an extension to Riemann integration, so we of course both suspected that they reduce to the same thing. Still, making the distinction does in fact matter - if we consider the expressions as Riemann integrals, it would not be valid to evaluate them using arguments based on monotone or dominated convergence.
I remember wondering why our lecturer never bothered to formally establish that Riemann integration is indeed a special case of Lebesgue integration. In this post, I hope to do exactly that. It's not quite as easy as you might have expected - there is a crucial subtlety involving null sets.
Recall the definition of Riemann integration, from Analysis I:
Let $f$ be a function of a closed interval, say $[0,1]$.
A partition $T$ of $[0,1]$ is an increasing sequence $0=t_0,t_1,\ldots,t_n=1$, which splits $[0,1]$ into $n$ intervals given by $[t_0,t_1]$, $[t_1,t_2]$, $\ldots$, $[t_{n-1}, t_n]$.
Define the lower sum over the partition $T$ to be $\displaystyle \sum_{i=0}^{n-1} \inf_{[t_i,t_{i+1}]} f$.
Define the upper sum over $T$ to be $\displaystyle \sum_{i=0}^{n-1} \sup_{[t_i,t_{i+1}]} f$.
$f$ is said to be integrable if, taking the supremum over the lower sums, and the infimum over the upper sums, the results both exist and coincide.
A partition $T$ of $[0,1]$ is an increasing sequence $0=t_0,t_1,\ldots,t_n=1$, which splits $[0,1]$ into $n$ intervals given by $[t_0,t_1]$, $[t_1,t_2]$, $\ldots$, $[t_{n-1}, t_n]$.
Define the lower sum over the partition $T$ to be $\displaystyle \sum_{i=0}^{n-1} \inf_{[t_i,t_{i+1}]} f$.
Define the upper sum over $T$ to be $\displaystyle \sum_{i=0}^{n-1} \sup_{[t_i,t_{i+1}]} f$.
$f$ is said to be integrable if, taking the supremum over the lower sums, and the infimum over the upper sums, the results both exist and coincide.
Interestingly, when researching this topic on the internet, I discovered that this definition of integration is actually usually called Darboux integration. What is usually called Riemann integration does not obtain upper and lower sums separately, but instead uses what wiki calls a tagged partition - given a partition, you also pick an arbitrary point $t_i$ inside each interval $[a_i,b_i]$ at which to evaluate the function, and then define the integral to be $\sum f(t_i)(b_i - a_i)$.
This is exactly how Dr. Dorrzapf defined Riemann integral when formally justifying more general constructions such as the path-integral in examples classes (although I vaguely remember that he arbitrarily set the tag $t_i$ to $a_i$).
The reason that our analysis lecturer could take such liberties with his impressionable Analysis students is that Darboux integration and Riemann integration are known to be equivalent. And, with the extra perspective gleaned from a beginner's course in measure theory, I think that his decision was fully justified - the upper and lower sum approach is much better preparation for the subsequent construction of Lebesgue integration.
Recall the definition of Lebesgue integration given in P&M. Our measure space is $([0,1], \mu)$, where $\mu$ is Lebesgue measure.
A simple function is a linear combination of indicator functions of measurable sets.
The integral $\mu(1_A)$ of the indicator function $1_A$ of a measurable set $A$ is defined to be $\mu(A)$. We extend this definition linearly to the set of simple functions.
Let $f$ be a non-negative, (Lebesgue) measurable function. The Lebesgue integral $\mu(f)$ is defined to be $\sup_G \mu(g)$, where $G$ is the set of simple functions dominated by $f$.
A general (measurable) function $f$ is said to be integrable if $\mu(|f|)$ is finite, in which case we define $\mu(f)=\mu(f^+)-\mu(f^-)$.
The integral $\mu(1_A)$ of the indicator function $1_A$ of a measurable set $A$ is defined to be $\mu(A)$. We extend this definition linearly to the set of simple functions.
Let $f$ be a non-negative, (Lebesgue) measurable function. The Lebesgue integral $\mu(f)$ is defined to be $\sup_G \mu(g)$, where $G$ is the set of simple functions dominated by $f$.
A general (measurable) function $f$ is said to be integrable if $\mu(|f|)$ is finite, in which case we define $\mu(f)=\mu(f^+)-\mu(f^-)$.
One slight quirk of notation - if $f$ is non-negative and measurable, $\mu(f)$ is defined whether or not $f$ is integrable.
The fundamental property of integration is the following:
If $f \leq g$ for any function $f$, $g$, then $\mu(f) \leq \mu(g)$.
The above property holds by construction for both Lebesgue and Riemann integrals.
Observation. A lower sum can be considered as a step function, i.e. a linear combination of indicator functions of (half-open) intervals.
Throughout the rest of this post, I will use the integral sign $\int$ to denote Riemann integration exclusively.
Corollary. If $f$ is Lebesgue measurable and Riemann integrable, then $f$ is Lebesgue integrable, and $\mu (f) = \int f \mathrm{d}x$.
Proof: $f$ is Riemann integrable, so there certainly exists one valid upper sum, meaning that $f$, and hence $|f|$, is bounded. Since we are in a finite measure space, $\mu(|f|) < \infty$, and so $f$ is Lebesgue integrable.
From the above observation, the set of lower sums can be considered as a subset of the set of simple functions. Let $G$ be the set of simple functions dominated by $f$, as before. Observe that every upper sum dominates $f$, and so every upper sum dominates every simple function in $G$. These two observations imply the following (you could worry about splitting $f$ up into positive and negative parts if you like, but there is zero difficulty involved):
$$\int f \mathrm{d}x = \sup\{\text{lower sums}\} \leq \sup_G \mu(g) = \mu(f) \leq \inf\{\text{upper sums}\} = \int f \mathrm{d}x $$
So $\int f \mathrm{d}x = \mu(f)$, as required. $\square$
From the above observation, the set of lower sums can be considered as a subset of the set of simple functions. Let $G$ be the set of simple functions dominated by $f$, as before. Observe that every upper sum dominates $f$, and so every upper sum dominates every simple function in $G$. These two observations imply the following (you could worry about splitting $f$ up into positive and negative parts if you like, but there is zero difficulty involved):
$$\int f \mathrm{d}x = \sup\{\text{lower sums}\} \leq \sup_G \mu(g) = \mu(f) \leq \inf\{\text{upper sums}\} = \int f \mathrm{d}x $$
So $\int f \mathrm{d}x = \mu(f)$, as required. $\square$
This is not quite sufficient to solve the problem as stated. To prove the full statement, we need to know that a Riemann integrable function is Lebesgue measurable. The proof of this result is actually more obscure than I thought it would be, drawing heavily from the properties of Lebesgue measure.
Recall the proof of Carathéodory's extension theorem, in which we construct the set $\mathcal{M}$ of outer-measurable sets. In fact, $M$ is the completion of the original $\sigma$-algebra, where by completion we mean that every subset of a set of zero measure is also measurable. Lebesgue measure is defined as the completion of the Borel measure on $\mathbb{R}$. This completeness property is precisely what we require.
Lemma. In a complete measure space, if $f$ is measurable, and $f=g$ almost everywhere, then $g$ is also measurable.
Proof: Write $\{f=g\}$ and $\{f \neq g\}$ for the sets on which $f=g$ and $f \neq g$ respectively. These are both measurable, and $\{f \neq g \}$ has measure zero, so every subset of it is measurable.
Given a measurable set $A \subseteq \mathbb{R}$, $g^{-1}(A)=(\left.g\right|_{f \neq g})^{-1}(A)\cup(\left.g\right|_{f = g})^{-1}(A)=(\left.g\right|_{f \neq g})^{-1}(A)\cup(\left.f\right|_{f = g})^{-1}(A)$. But $(\left.g\right|_{f \neq g})^{-1}(A)$ is a subset of the null set $\{f \neq g\}$, so is measurable, and $(\left.f\right|_{f = g})^{-1}(A)$ is also measurable.
Hence $g$ is measurable. $\square$
Given a measurable set $A \subseteq \mathbb{R}$, $g^{-1}(A)=(\left.g\right|_{f \neq g})^{-1}(A)\cup(\left.g\right|_{f = g})^{-1}(A)=(\left.g\right|_{f \neq g})^{-1}(A)\cup(\left.f\right|_{f = g})^{-1}(A)$. But $(\left.g\right|_{f \neq g})^{-1}(A)$ is a subset of the null set $\{f \neq g\}$, so is measurable, and $(\left.f\right|_{f = g})^{-1}(A)$ is also measurable.
Hence $g$ is measurable. $\square$
In a non-complete measure space, the above lemma does not hold. If there exists a non-measurable set $Y$ that is a subset of a null set, then for any measurable function $f$, $g \equiv f + \chi_Y$ is equal to $f$ almost everywhere, but is not measurable.
In order to finish off, I'm going to use heavy machinery that I'm not going to prove (it's sufficiently interesting that I plan to revisit it in future, however).
Theorem. (Riemann-Lebesgue) A function $f$ is Riemann integrable $\Leftrightarrow$ $f$ is bounded and continuous almost everywhere.
Corollary. If $f$ is Riemann integrable, then $f$ is (Lebesgue) measurable.
Proof: By the theorem, $f$ is equal almost everywhere to the function $g$ which is continuous on a measurable set, and zero on its complement. $g$ is clearly measurable, so using completeness of Lebesgue measure, $f$ is also measurable. $\square$
Conclusion. If $f$ is properly Riemann integrable, then $f$ is Lebesgue integrable, and the Riemann and Lebesgue integrals coincide.
Remark: from the discussion above, it is clear that there can exist a function that is Riemann integrable but not Borel measurable.
There are a few more things to say about the link between Riemann integration and Lebesgue integration. Tomorrow, I will post a short entry discussing examples of non-Riemann integrable functions that are Lebesgue integrable, as well as making a few remarks about improper Riemann integrals.
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