Advanced Analysis

Properties of the Riemann Integral

Let \(f\) be a bounded real-valued function on \([a,b]\). If \(U(f,P)\) and \(L(f,P)\) denote upper and lower sums for a given partition \(P\) of \([a,b]\), then we say that \(f\) is Riemann integrable on \([a,b]\) when
\[ \sup_{P} L(f,P) = \inf_P U(f,P) \]
and define
\[ \int_a^b f(t) dt := \inf_P U(f,P). \]

1. Algebraic Properties of the Integral

Proposition (Linearity)
If \(f\) and \(g\) are Riemann-integrable functions on \([a,b]\) and if \(c \in {\mathbb R}\), then \(c f + g\) is Riemann integrable on \([a,b]\) and
\[{}\int_a^b (c f + g)(t) dt{}\]
\[{}= c \int_a^b f(t) dt{}\]
\[{}+ \int_a^b g(t) dt.{}\]
Proof
Meta (Main Idea)
Without loss of generality, \(c \neq 0\). Boundedenss of \(f\) and \(g\) implies boundedness of \(c f + g\). Using a common refinement, there exists a partition \(P\) such that
\[ U(f,P) - L (f,P) < \frac{\epsilon}{2|c|}\]
and
\[ U(g,P) - L (g,P) < \frac{\epsilon}{2}. \]
Now if \(c > 0\),
\[{}\sup_{x \in I} (c f(x) + g(x)){}\]
\[{}\leq c \sup_{x \in I} f(x){}\]
\[{}+ \sup_{x \in I} g(x) {}\]
and
\[{}\inf_{x \in I} (c f(x) + g(x)){}\]
\[{}\geq c \inf_{x \in I} f(x){}\]
\[{}+ \inf_{x \in I} g(x).{}\]
Now multiplying by \(|I|\) and summing over \(I \in P\) gives that
\[{}c L(f,P) + L(g,P){}\]
\[{}\leq L(cf + g,P){}\]
\[{}\leq U(cf+g,P){}\]
\[{}\leq c U(f,P) + U(g,p).{}\]
By choice of \(P\), the quantities on the left and right sides of this chain of inequalities differ from
\[ c \int_a^b f(t) dt + \int_a^b g(t) dt \]
by at most \(\epsilon\). Thus it follows that
\[{}\inf_P U(f,P){}\]
\[{}\leq \epsilon + c \int_a^b f(t) dt {}\]
\[{}+ \int_a^b g(t) dt{}\]
and
\[{}\sup_P L(f,P){}\]
\[{}\geq -\epsilon + c \int_a^b f(t) dt{}\]
\[{}+ \int_a^b g(t) dt{}\]
and since \(\epsilon\) is arbitrary,
\[{}\inf_P U(f,P){}\]
\[{}= \sup_P L(f,P){}\]
\[{}= c \int_a^b f(t) dt{}\]
\[{}+ \int_a^b g(t) dt.{}\]
If \(c < 0\), a modification is necessary to reflect the fact that multiplication by \(c\) reverses ordering.
Corollary
If \(f\) and \(g\) are Riemann integrable on \([a,b]\) and \(f(x) \leq g(x)\) for all \(x \in [a,b]\), then
\[ \int_a^b f(t) dt \leq \int_a^b g(t) dt. \]
Proof
Meta (Main Idea)
It suffices to know that \(g-f\) is Riemann integrable and has nonnegative integral. That's easy to see by showing that each lower sum must be nonnegative.
Corollary (Triangle Inequality)
If \(f\) is Riemann integrable on \([a,b]\) then \(|f(x)|\) is as well and
\[ \left| \int_a^b f(t) dt \right| \leq \int_a^b |f(t)| dt. \]
Proof
Meta (Main Idea)
To prove integrability, show that
\[{}\sup_{x \in I} |f(x)| - \inf_{y \in I} |f(y)|{}\]
\[{}\leq \sup_{x \in I} f(x) - \inf_{y \in I} f(y).{}\]
To prove the inequality, use the fact that \(- |f(x)| \leq f(x) \leq |f(x)|\) and use the previous corollary.
Proposition (Subdivision)
If \(f\) is bounded and Riemann integrable on \([a,b]\) and if \(c \in (a,b)\), then \(f\) is Riemann integrable on \([a,c]\) and \([c,b]\) and
\[ \int_a^b f(t) dt = \int_a^c f(t) dt + \int_c^b f(t) dt. \]
Proof
Meta (Main Idea)
Let \(P\) be any partition of \([a,b]\) such that \(U(f,P) - L(f,P) < \epsilon\). If the point \(c\) is not an endpoint of intervals \(I \in P\), subdivide the unique interval \(I' := [a',b']\) in \(P\) which contains \(c\) into \([a',c]\) and \([c,b']\). Subdivisions never increase the value of upper sums, nor do they decrease the value of lower sums, so without loss of generality, it may be assumed that the intervals in \(P\) are each contained in either \([a,c]\) or \([c,b]\). If \(P_L\) are the intervals of \(P\) which are contained in \([a,c]\) and \(P_R\) are the intervals of \(P\) which are contained in \([c,b]\), then \(P_L\) will necessarily be a partition of \([a,c]\) and \(P_R\) will partition \([c,b]\)
(because the intervals are themselves nonoverlapping and cover all of \([a,b]\).) Then
\[ U(f,P) = U(f,P_L) + U(f,P_R) \]
and
\[ L(f,P) = L(f,P_L) + L(f,P_R) \]
so
\[{}U(f,P_L) - L(f,P_L){}\]
\[{}+ U(f,P_R) - L(f,P_R) < \epsilon.{}\]
This forces integrability on both \([a,c]\) and \([c,b]\). But now we can take \(\epsilon \rightarrow 0\) and the identity \(U(f,P) = U(f,P_L) + U(f,P_R)\) to conclude the additivity of the integral (because each upper sum differs from the corresponding integral by at most \(\epsilon\)).

2. Integrability of Pointwise Products

Integrability of pointwise products \(fg\) is a surprisingly subtle question because there are no algebraic identities which dictate what the value of the integral should be. To prove this result, it's most straightforward to show first that squaring preserves integrability and then use algebraic laws to extend this result.
Proposition
Suppse that \(f\) is a bounded function on some closed, bounded interval \(I \subset {\mathbb R}\). Then
\[{} \sup_{x \in I} (f(x))^2 - \inf_{y \in I} (f(y))^2{}\]
\[{}\leq 2 \sup_{z \in I} |f(z)| \left( \sup_{x \in I} f(x) - \inf_{y \in I} f(y) \right).{}\]
Proof
For any \(\epsilon > 0\), we know there is \(x_\star \in I\) such that
\[ (f(x_\star))^2 \geq -\epsilon + \sup_{x \in I} (f(x))^2 \]
and \(y \in I\) such that
\[ (f(y_\star))^2 \leq \epsilon + \inf_{y \in I} (f(y))^2. \]
Thus
\[{}\sup_{x \in I} (f(x))^2 - \inf_{y \in I} (f(y))^2{}\]
\[{}\leq 2 \epsilon + (f(x_\star))^2 - (f(y_\star))^2.{}\]
Without loss of generality, it may be assumed that \((f(x_\star))^2 - (f(y_\star))^2 \geq 0\), since otherwise one could increase \((f(x_\star))^2\) by simply choosing \(x_\star = y_\star\). The difference of squares factorization gives that
\[{}(f(x_\star))^2 - (f(y_\star))^2{}\]
\[{}= (f(x_\star) + f(y_\star)){}\]
\[{}\cdot (f(x_\star) - f(y_\star)).{}\]
Since the product is nonnegative, both terms must be nonnegative or nonpositive. In the former case, we can use
\[ f(x_\star) + f(y_\star) \leq 2 \sup_{z \in I} |f(z)| \]
and
\[ f(x_\star) - f(y_\star) \leq \sup_{x \in I} f(x) - \inf_{y \in I} f(y) \]
to conclude that
\[{}\sup_{x \in I} (f(x))^2 - \inf_{y \in I} (f(y))^2{}\]
\[{}\leq 2 \sup_{z \in I} |f(z)| \left( \sup_{x \in I} f(x) - \inf_{y \in I} f(y) \right){}\]
\[{}+ 2 \epsilon.{}\]
In the remaining case, we have symmetry which allows us to replace \(f\) by \(-f\) to conclude that
\[{}\sup_{x \in I} (f(x))^2 - \inf_{y \in I} (f(y))^2{}\]
\[{}\leq 2 \sup_{z \in I} |-f(z)| {}\]
\[{}\cdot \left( \sup_{x \in I} \left[ -f(x) \right] - \inf_{y \in I} \left[ -f(y) \right] \right){}\]
\[{}+ 2 \epsilon.{}\]
In particular, the extra minus signs on the right-hand side do not change its value, and we conclude that
\[{}\sup_{x \in I} (f(x))^2 - \inf_{y \in I} (f(y))^2{}\]
\[{}\leq 2 \sup_{z \in I} |f(z)| \left( \sup_{x \in I} f(x) - \inf_{y \in I} f(y) \right){}\]
\[{}+ 2 \epsilon{}\]
for any \(\epsilon > 0\). Letting \(\epsilon \rightarrow 0^+\) establishes the desired inequality.
Theorem
If \(f\) is Riemann integrable on \([a,b]\), then the pointwise square \(f^2\) is also Riemann integrable.
Proof
Riemann integrability implies boundedness; let \(M > 0\) be an upper bound for \(|f|\) on \([a,b]\). The Proposition above implies that
\[{}U(f^2,P) - L(f^2,P){}\]
\[{}\leq 2 M \left[ U(f,P) - L(f,P) \right]{}\]
for any partition \(P\) of \([a,b]\). Since the right-hand side can be made as small as desired for appropriate choice of \(P\), the left-hand side must also be as small as desired for appropriate \(P\).
Corollary
If \(f\) and \(g\) are Riemann integrable functions on \([a,b]\), then the pointwise product \(fg\) is also Riemann integrable.
Proof
Use the identity \(fg = \frac{1}{4} ((f+g)^2-(f-g)^2)\), the Riemann integrability of squares, and the fact that Riemann integrable functions are a real vector space to make the desired conclusion.

3. Integrals and Uniform Convergence

Theorem
Suppose that \(\{f_n\}_{n=1}^\infty\) is a sequence of Riemann integrable functions on \([a,b]\) which converge uniformly to some limit function \(f\). Then \(f\) must be Riemann integrable as well and the limit passes through the integral, i.e.,
\[ \lim_{n \rightarrow \infty} \int_a^b f_n(t) dt = \int_a^b f(t) dt. \]
Proof
First, observe that there is some \(N\) such that \(|f_n(x) - f(x)| \leq 1\) for all \(n \geq N\) and all \(x \in [a,b]\). Since \(f_N\) is bounded and \(f\) differs from \(f_N\) pointwise by at most \(1\), \(f\) must also be bounded. In general, if \(\epsilon > 0\) is fixed and \(|f_n(x) - f(x)| < \epsilon/(2 + 2|b-a|)\) for all \(n \geq N\) and all \(x \in [a,b]\), then it follows that for any \(I\) in a partition \(P\) of \([a,b]\),
\[{}|I| \sup_{x \in I} f(x){}\]
\[{}\leq \frac{\epsilon}{2 + 2|b-a|} |I| + |I| \sup_{x \in I} f_n(x){}\]
with an analogous inequality for the infimum. Thus
\[{}-\frac{\epsilon}{2+2|b-a|} (b-a) + L(f_n,P){}\]
\[{}\leq L(f,P) \leq U(f,P){}\]
\[{}\leq \frac{\epsilon}{2+2|b-a|}(b-a){}\]
\[{}+ U(f_n,P){}\]
for all \(n \geq N\). Fix any such \(n\) and choose \(P\) appropriately so that both \(U(f_n,P)\) and \(L(f_n,P)\) differ from the integral of \(f_n\) by at most \(\epsilon/2\). It follows then that both \(L(f,P)\) and \(U(f,P)\) differ from
\[ \int_a^b f_n(t) dt \]
by at most \(\epsilon\). There are two important observations:
  1. Since \(U(f,P)\) and \(L(f,P)\) belong to an interval of real numbers which has length at most \(2 \epsilon\) and \(\epsilon\) is arbitrary, it follows that the supremum of upper sums and infimum of lower sums must be equal and that \(f\) must be Riemann integrable. Consequently
    \[ \int_a^b f(t) dt \]
    must itself differ from \(\int_a^b f_n(t) dt\) by at most \(\epsilon\) for all \(n \geq N\).
  2. By the previous part, we also have that the limit of \(\int_a^b f_n(t) dt\) exists as \(n \rightarrow \infty\) and is simply equal to the integral of \(f\).
Exercises
  1. Show that even when it is known that the integrals of both \(f\) and \(g\) are equal to \(1\) on the interval \([0,1]\), there is no limit to how large \(\int_0^1 f(t) g(t) dt\) could be.
    Hint
    Take \(f\) and \(g\) to be equal to large constant values on some short interval \([0,\delta]\) and zero elsewhere.
  2. Give a proof of the fact that all continuous functions on \([a,b]\) are Riemann integrable by using the results discussed on this page combined with the fact that \(f(x) := x\) is Riemann integrable on \([a,b]\) together with the Weierstrass Approximation Theorem.