Riemann Integration I
1. Partitions
Definition
Given an interval \([a,b]\), a partition \(P\) of \([a,b]\) is a finite set of nonoverlapping intervals \(P := \{I_1,\ldots,I_N\}\) whose union is \([a,b]\). In other words,
- Any two intervals \(I, I' \in P\) which are not equal intersect at most at their endpoints only, and
- The union of all intervals in \(P\) is \([a,b]\).
When the intervals \(I_1,\ldots,I_N\) of a partition \(P\) are arranged in order based on their lowest endpoint, the partition will always have the form
\[ P = \{ [a,x_1],[x_1,x_2],\ldots,[x_{N-1},b] \}\]
for some points \(x_1 < x_2 < \cdots < x_{N-1}\) in \((a,b)\). The proof of this fact can be established by induction on the number of intervals in the partition:
- Any partition of \([a,b]\) which consists of a single interval must have the form \(\{[a,b]\}\) because it has to cover and be contained in \([a,b]\).
- If the structure of all partitions of cardinality \(N\) is known to be as described above, let \(P\) be a partition of \([a,b]\) which has cardinality \(N+1\). There must be some interval \(I_{N+1}\) containing the point \(b\). Let \(x_{N}\) denote the left endpoint of \(I_{N+1}\). Because \(I_1,\ldots,I_N\) don't overlap \(I_{N+1}\), their union must belong to \([a,x_{N}]\). But since \(I_1 \cup \cdots \cup I_{N+1} = [a,b]\), it follows that \([a,x_{N}) \subset I_1 \cup \cdots \cup I_{N}\). However, since \(I_1 \cup \cdots \cup I_N\) must also be closed, it must at least contain the closure of \([a,x_N)\), which is exactly \([a,x_N]\). Thus \(I_1,\ldots,I_N\) are nonoverlapping and have union \([a,x_N]\), meaning that they form a partition \(P'\) of \([a,x_N]\). The induction hypothesis then guarantees that after suitably reordering \(I_1,\ldots,I_N\), they have the form \([a,x_1],\ldots,[x_{N-1},x_N]\). Since \(I_N = [x_N,b]\), \(P\) itself also has the claimed structure.
Definition
Given two partitions \(P\) and \(P'\) of the interval \([a,b]\), we say that \(P'\) is a refinement of \(P\) when every interval of \(I' \in P'\) is entirely contained in some \(I \in P\).
Suppose that
\[ P = \{ [a,x_1],[x_1,x_2],\ldots,[x_{N-1},b] \}, \]
\[ P' = \{ [a,y_1],[y_1,y_2],\ldots,[y_{M-1},b] \}, \]
and \(P'\) is a refinement of \(P\). We claim that every endpoint \(x_1,\ldots,x_{N-1}\) of an interval of \(P\) must equal \(y_i\) for some \(i\); in other words, we claim that the set of intermediate endpoints of \(P'\) must contain all intermediate endpoints of \(P\). This is because each \(x_i\) must belong to some interval \(I'\) of \(P'\), but can't belong to its interior, as then \(I'\) would not be entirely contained in any one interval \(I \in P\). So a partition effectively subdivides (and indeed partitions!) intervals in \(P\) into smaller ones.
Proposition
Given partitions \(P_1\) and \(P_2\) of \([a,b]\), there is a partition \(P_3\) of \([a,b]\) which is a refinement of both \(P_1\) and \(P_2\). It is called a common refinement.
Proof
Let \(S_1\) be the set of all endpoints of intervals in \(P_1\) and let \(S_2\) be the set of all endpoints of intervals in \(P_2\). Let \(S = S_1 \cup S_2\). We know that \(a\) and \(b\) have to belong to both \(S_1\) and \(S_2\), so we can write \(S = \{a,y_1,\ldots,y_{M-1},b\}\) for some points \(y_1 < \ldots < y_{M-1} \in (a,b)\). Let \(P_3\) be the partition \(\{[a,y_1],\ldots,[y_{M-1},b]\}\). Every interval in \(P_3\) must be entirely contained in some element of \(P_1\). To see this, let \(z\) be the midpoint of \(I_3 \in P_3\) and let \(I_1 \in P_1\) contain \(z\). By construction, if \(I_3\) equals \([y,y']\) for some \(y,y'\), then \(y\) is the largest point in \(S\) which is less than or equal to \(z\) and \(y'\) is the smallest point in \(S\) which is greater than or equal to \(z\). If \(I_1 := [x,x']\), then both \(x\) and \(x'\) belong to \(S\) and clearly \(x \leq z \leq x'\) because \(z \in I_1\). Because \(y\) and \(y'\) are extremal, \(x \leq y\) and \(y' \leq x'\), so \(I_3 \subset I_1\). A symmetric argument establishes that \(P_3\) is a refinement of \(I_2\).
2. Upper and Lower Sums
Definition
Given any bounded function \(f : [a,b] \rightarrow {\mathbb R}\) and any partition \(P\) of \([a,b]\), we define the upper and lower sums of \(f\) with respect to \(P\) by the formulas
\[ U(f,P) := \sum_{I \in P} |I| \sup_{x \in I} f(x), \]
\[ L(f,P) := \sum_{I \in P} |I| \inf_{x \in I} f(x), \]
respectively, where \(|I|\) denotes the length of the interval \(I\).
Proposition
For any bounded function \(f\) on \([a,b]\),
- For any partition \(P\) of \([a,b]\),\[ L(f,P) \leq U(f,P). \]
- For any partition \(P\) of \([a,b]\) and any refinement \(P'\) of \(P\),\[{}L(f,P){}\]\[{}\leq L(f,P'){}\]\[{}\leq U(f,P'){}\]\[{}\leq U(f,P).{}\]
- For any partitions \(P_1\) and \(P_2\) of \([a,b]\),\[ L(f,P_1) \leq U(f,P_2). \]
Proof
- This follows immediately from the fact that \(\inf_{x \in I} f(x) \leq \sup_{x \in I} f(x)\) for each \(I \in P\).
- For any interval \(I \in P\), the intervals \(I' \in P'\) which are contained in \(I\) form a partition of \(I\). Thus\[{}\sum_{\substack{I' \in P' \\ I' \subset I}} |I'| \sup_{x \in I'} f(x){}\]\[{}\leq \sum_{\substack{I' \in P' \\ I' \subset I}} |I'| \sup_{x \in I} f(x){}\]\[{}\leq \sup_{x \in I} f(x) \sum_{\substack{I' \in P' \\ I' \subset I}} |I'|{}\]\[{}= |I| \sup_{x \in I} f(x).{}\]Now sum both sides over all \(I \in P\) to conclude that \(U(f,P') \leq U(f,P)\). A symmetric argument also establishes that \(L(f,P') \geq L(f,P)\).
- Let \(P'\) be a common refinement of \(P_1\) and \(P_2\). Then\[{}L(f,P_1){}\]\[{}\leq L(f,P'){}\]\[{}\leq U(f,P'){}\]\[{}\leq U(f,P_2).{}\]
Corollary
For any bounded function \(f : [a,b] \rightarrow {\mathbb R}\),
\[{}\sup_{P \text{ partitions } [a,b]} L(f,P){}\]
\[{}\leq \inf_{P \text{ partitions } [a,b]} U(f,P).{}\]
We call the quantity on the left-hand side the lower integral and denote it
\[ (L)\int_a^b f(t) dt. \]
The quantity on the right-hand side is called the upper integral and denoted
\[ (U) \int_a^b f(t) dt. \]
Proof
Let
\[ E := \{ L(f,P) \ : \ P \text{ partitions } [a,b] \}\]
and
\[ F := \{ U(f,P) \ : \ P \text{ partitions } [a,b] \}. \]
Both sets are nonempty, and for any partition \(Q\) of \([a,b]\) we have by the previous proposition that \(U(f,Q)\) is an upper bound for \(E\) and \(L(f,Q)\) is a lower bound for \(F\). Thus \(E\) has an supremum and \(F\) has an infimum. By defintion,
\[ \sup E := (L) \int_a^b f(t) dt \]
and
\[ \inf F := (U) \int_a^b f(t) dt. \]
Because the supremum is the least upper bound, \(\sup E \leq U(f,Q)\) for all partitions \(Q\) of \([a,b]\). Likewise \(\inf F \geq L(f,Q)\) for all partitions \(Q\) of \([a,b]\). Now taking the infimum and supremum over \(Q\) gives that \(\sup E \leq \inf_Q U(f,Q) = \inf F\) as desired.
Definition
We say that a bounded function \(f : [a,b] \rightarrow {\mathbb R}\) is Riemann integrable on \([a,b]\) when
\[ (L) \int_a^b f(t) dt = (U) \int_a^b f(t) dt. \]
When such a function is integrable, we define
\[ \int_a^b f(t) dt := (U) \int_a^b f(t) dt. \]
Example
The function \(f : [0,1] \rightarrow {\mathbb R}\) which equals \(1\) at all points \(x \in {\mathbb{Q}}\) and \(0\) elsewhere is not Riemann integrable on \([0,1]\).
Proof
For any interval \(I\), \(\sup_{x\in I} f(x) = 1\) because \(I\) contains rationals. Also \(\inf_{x \in I} f(x) = 0\) because \(I\) contains irrationals. Thus
\[ U(f,P) = \sum_{I \in P} |I| \sup_{x \in I} f(x) = \sum_{I \in P} |I| = 1 \]
and
\[ L(f,P) = \sum_{I \in P} |I| \inf_{x \in I} f(x) = \sum_{I \in P} 0 = 0. \]
So the lower integral must be zero and the upper integral must be \(1\).
Example
The function \(f(x) = x\) is Riemann integrable on \([0,1]\).
Proof
Fix some natural number \(n\) and consider the partition
\[{}P{}\]
\[{}:= \{[0,1/n],{}\]
\[{}[1/n,2/n],\ldots,{}\]
\[{}[(n-1)/n,1]\}.{}\]
We have that
\[{}U(f,P){}\]
\[{}= \sum_{i=0}^{n-1} \frac{1}{n} \sup_{x \in [i/n,(i+1)/n]} x{}\]
\[{}= \sum_{i=0}^{n-1} \frac{1}{n} \frac{i+1}{n}{}\]
and similarly that
\[ L(f,P) = \sum_{i=0}^{n-1} \frac{1}{n} \frac{i}{n}. \]
Thus
\[{}U(f,P) - L(f,P){}\]
\[{}= \sum_{i=0}^{n-1} \frac{1}{n} \left( \frac{i+1}{n} - \frac{i}{n} \right){}\]
\[{}= \sum_{i=0}^{n-1} \frac{1}{n^2} = \frac{1}{n}.{}\]
Since
\[ (U) \int_0^1 x dx \leq U(f,P) \]
and
\[ (L) \int_0^1 x dx \geq L(f,P) \]
for any partition \(P\), it follows that
\[{}0{}\]
\[{}\leq (U) \int_0^1 x dx - (L) \int_0^1 x dx{}\]
\[{}\leq U(f,P) - L(f,P) = \frac{1}{n}.{}\]
Since \(n\) is arbitrary, the upper and lower integrals must be equal.
Exercises
- Show that a bounded function \(f : [a,b] \rightarrow {\mathbb R}\) is Riemann integrable on \([a,b]\) if and only if for every \(\epsilon > 0\), there is some partition \(P\) such that \(U(f,P) - L(f,P) < \epsilon\).HintWhen \(f\) is assumed to be integrable, take a partition \(P_1\) such that \(U(f,P_1)\) is close to the upper integral, a partition \(P_2\) such that \(L(f,P_2)\) is close to the lower integral, and then study the properties of a common refinement \(P\) of \(P_1\) and \(P_2\).
- If \(f\) and \(g\) are bounded functions on \([a,b]\) and \(f(x) \leq g(x)\) for all \(x \in [a,b]\), show that\[ (L) \int_a^b f(t) dt \leq (L) \int_a^b g(t) dt, \]\[ (U) \int_a^b f(t) dt \leq (U) \int_a^b g(t) dt. \]Do not assume that \(f\) and \(g\) are Riemann integrable.