Real Analysis: Integrals

Problems

1. Boundedness: Show that if $f \in \mathscr{R}([a, b]),$ then there exists some $M > 0$ such that

   $$\int\limits_a^b f(x) \, dx \leq M(b-a).$$

   Because $f \in \mathscr{R}([a, b]),$ $f$ is bounded. Therefore there exists some $M > 0$ such that $|f(x)| < M$ for all $x \in [a, b].$ Let $P$ be a finite interval subdivision of $[a, b].$ The upper bound of $U(f, P)$ on the integral, and the subsequent upper bound of $M(a, b)$ on that, shows the desired result:

   $ \displaystyle \int\limits_a^b f(x) \, dx \leq U(f, P) \\ \displaystyle \int\limits_a^b f(x) \, dx \leq \sum\limits_{p_i \in P} \sup(f(p_i))\mu(p_i) \\ \displaystyle \int\limits_a^b f(x) \, dx \leq \sum\limits_{p_i \in P} M \mu(p_i) \\ \displaystyle \int\limits_a^b f(x) \, dx \leq M(b-a). $

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2. Continuous compositions of integrable functions are integrable: Show that if $f$ is a Riemann integrable function defined on $[a, b]$ and bounded by $[A, B]$ and if $g$ is a continuous function defined on $[A, B]$ that $g \circ f$ is integrable on $[a, b].$

   Because $g$ is continuous and $[A, B]$ is compact, $g$ is uniformly continuous. Therefore for every $\varepsilon > 0$, there is a $\delta > 0$ where such that $|g(x) - g(y)| < \varepsilon$ whenever $|x - y| < \delta.$ As a further condition, we select $\delta < \varepsilon.$

   Because $f$ is Riemann integrable on $[a, b],$ there is a partition $P = \{p_1, \ldots, p_n\}$ such that $U(f, P) - L(f, P) < \delta^2.$

   Partition P into two sets:

   $A = \{ a_i = p_i \in P : \sup(f(p_i)) - \inf(f(p_i)) < \delta \} \\ B = \{ b_i = p_i \in P : \sup(f(p_i)) - \inf(f(p_i)) \geq \delta \}$

   Plainly, $A$ is those intervals in P whose diameter under $f$ is less than $\delta,$ and $B$ is those intervals in P whose diameter under f is greater than or equal to $\delta.$

   For $a_i \in A,$ it follows that $\sup(g(f(a_i))) - \inf(g(f(a_i))) \leq \varepsilon.$

   For $b_i \in B,$ it follows that $\sup(g(f(b_i))) - \inf(g(f(b_i))) \leq \text{Diam}(g([A,B])).$

   By the construction of $B,$ we derive the following inequality:

   $\sum\limits_{b_i \in B} \delta \mu(b_i) \leq \sum\limits_{b_i \in B} \left(\sup(f(b_i)) - \inf(f(b_i))\right) \mu(b_i) $

   Since $P = B \cup A,$ the righthand side is less than the difference of the upper and lower Darboux sums:

   $\sum\limits_{b_i \in B} \delta \mu(b_i) \leq U(f, P) - L(f, P).$

   Our choice of partition $P$ for $f$ lets us then take the following step:

   $ \sum\limits_{b_i \in B} \delta\mu(b_i) < \delta^2 \\ \sum\limits_{b_i \in B} \mu(b_i) < \delta$

   From this we can recombine the two sums of partition elements in $A$ and $B$ to arrive at the desired conclusion:

   $ U(g\circ f, P) - L(g \circ f, P) = \sum\limits_{a_i \in A} (\sup(g(f(a_i)) - \inf(g(f(p_i))))\mu(a_i) + \sum\limits_{b_i \in B} (\sup(g(f(b_i)) - \inf(g(f(b_i))))\mu(b_i) \\ U(g\circ f, P) - L(g \circ f, P) \leq \varepsilon\mu(P) + \text{Diam}(g([A,B]))\delta \\ U(g\circ f, P) - L(g \circ f, P) < \varepsilon(\mu(P) + \text{Diam}(g([A,B]))) $

   Since $\varepsilon$ is arbitrary, it follows that $g \circ f$ is Riemann integrable.

   Show Answer

3. Show that if $c \in \mathbb{R}$ and $f \in \mathscr{R}([a, b]),$ then $cf \in \mathscr{R}([a, b]).$

   If $c = 0,$ then $cf = 0$ and is Riemann integrable.

   Consider the case when $c > 0.$ Since $f \in \mathscr{R}([a, b]),$ for each $\frac{\varepsilon}{c} > 0$ there exists a partition $P$ such that $U(f, P) - L(f, P) < \frac{\varepsilon}{c}.$ Expanding the definition of $L(f, P)$ shows that

   $ L(cf, P) = \sum\limits_{p_i \in P} \inf(cf(p_i))\mu(p_i) \\ L(cf, P) = \sum\limits_{p_i \in P} c\inf(f(p_i))\mu(p_i) \\ L(cf, P) = c\sum\limits_{p_i \in P} \inf(f(p_i))\mu(p_i) \\ L(cf, P) = cL(f, P) $

   The same reasoning shows that $U(cf, P) = cU(f, P).$ Therefore

   $ U(cf, P) - L(cf, P) = cU(f, P) - cL(f, P) \\ U(cf, P) - L(cf, P) = c(U(f, P) - L(f, P)) \\ U(cf, P) - L(cf, P) < c\left(\dfrac{\varepsilon}{c}\right) \\ U(cf, P) - L(cf, P) < \varepsilon $

   Therefore $cf \in \mathscr{R}([a, b]).$

   Consider the alternate case when $c < 0.$ Since $f \in \mathscr{R}([a, b]),$ for each $\frac{\varepsilon}{-c} > 0$ there exists a partition $P$ such that $U(f, P) - L(f, P) < \frac{\varepsilon}{-c}.$ Expanding the definition of $L(f, P)$ shows that

   $ L(cf, P) = \sum\limits_{p_i \in P} \inf(cf(p_i))\mu(p_i) \\ L(cf, P) = \sum\limits_{p_i \in P} c\sup(f(p_i))\mu(p_i) \\ L(cf, P) = c\sum\limits_{p_i \in P} \sup(f(p_i))\mu(p_i) \\ L(cf, P) = cU(f, P) $

   The same reasoning shows that $U(cf, P) = cL(f, P).$ Therefore

   $ U(cf, P) - L(cf, P) = cL(f, P) - cU(f, P) \\ U(cf, P) - L(cf, P) = c(L(f, P) - U(f, P)) \\ U(cf, P) - L(cf, P) = -c(U(f, P) - L(f, P)) \\ U(cf, P) - L(cf, P) < c\left(\dfrac{\varepsilon}{-c}\right) \\ U(cf, P) - L(cf, P) < \varepsilon $

   Therefore $cf \in \mathscr{R}([a, b]).$

   Show Answer

4. Homogeneity: Show that if $f \in \mathscr{R}([a, b])$ and $c \in \mathbb{R},$ then $\displaystyle \int\limits_a^b cf(x) \, dx = c\int\limits_a^b f(x) \, dx.$

   As shown in the prior proof, $cf \in \mathscr{R}([a, b]).$

   If $c = 0,$ then $cf = 0,$ and the integral of both over $[a, b]$ is $0.$

   Consider the case where $c > 0.$ For any finite interval subdivision $P,$ note that $L(cf, P) = cL(f, P)$ and $U(cf, P) = cU(f, P).$ It follows that $L^*(cf, [a, b]) = cL^*(f, [a, b])$ and $U^*(cf, [a, b]) = cU^*(f, [a, b]).$ It follows from the definition of the Darboux integral that

   $$\int\limits_a^b cf(x) \, dx = c\int\limits_a^b f(x) \,dx.$$

   Consider the case where $c < 0.$ For any finite interval subdivision $P,$ note that $L(cf, P) = cU(f, P)$ and $U(cf, P) = cL(f, P).$ It follows that $L^*(cf, [a, b]) = cU^*(f, [a, b])$ and $U^*(cf, [a, b]) = cL^*(f, [a, b]).$ It follows from the definition of the Darboux integral that

   $$\int\limits_a^b cf(x) \, dx = c\int\limits_a^b f(x) \,dx.$$

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5. Show that if $f, g \in \mathscr{R}([a, b]),$ then $f + g \in \mathscr{R}([a, b]).$

   Because $f, g \in \mathscr{R}([a, b]),$ for every $\frac{\varepsilon}{2} > 0$ there exist finite interval subdivisions $P_f$ and $P_g$ such that $U(f, P_f) - L(f, P_f) < \frac{\varepsilon}{2}$ and $U(g, P_g) - L(g, P_g) < \frac{\varepsilon}{2}.$ Set $P = P_f \# P_g.$ Since $P$ is finer than both $P_f$ and $P_g,$ it follows that $U(f, P) - L(f, P) < \dfrac{\varepsilon}{2}$ and $U(g, P) - L(g, P) < \frac{\varepsilon}{2}.$

   From the definition of lower Darboux sum, we see that

   $ L(f, P) + L(g, P) = \sum\limits_{p_i \in P} \inf(f(p_i))\mu(p_i) + \sum\limits_{p_i \in P} \inf(g(p_i))\mu(p_i) \\ L(f, P) + L(g, P) = \sum\limits_{p_i \in P} \left(\inf(f(p_i)) + \inf(g(p_i))\right)\mu(p_i) \\ L(f, P) + L(g, P) \leq \sum\limits_{p_i \in P} \left(\inf(f(p_i) + g(p_i))\right)\mu(p_i) \\ L(f, P) + L(g, P) \leq L(f + g, P) $

   The same logic shows that $U(f + g, P) \leq U(f, P) + U(g, P)$

   Combining these two inequalities shows us that

   $ U(f + g, P) - L(f + g, P) \leq U(f, P) + U(g, P) - L(f, P) - L(g, P) \\ U(f + g, P) - L(f + g, P) \leq U(f, P) - L(f, P) + U(g, P) - L(g, P) \\ U(f + g, P) - L(f + g, P) < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} \\ U(f + g, P) - L(f + g, P) < \varepsilon $

   Therefore $f + g \in \mathscr{R}([a, b]).$

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6. Additivity: Show that if $f, g \in \mathscr{R}([a, b]),$ then $\displaystyle \int\limits_a^b f(x) + g(x) \, dx = \int\limits_a^b f(x) \, dx + \int\limits_a^b g(x) \, dx.$

   To prove the claim, we show that the integral of the sum is bounded both above and below by the sum of the integrals.

   Part 1:

   As shown in the prior proof, $f + g \in \mathscr{R}([a, b]).$ Likewise, for any $\frac{\varepsilon}{2} > 0$ there exists a finite interval subdivision $P$ such that

   $$U(f, P) - L(f, P) < \dfrac{\varepsilon}{2} \text{ and } U(g, P) - L(g, P) < \dfrac{\varepsilon}{2}.$$

   It follows from the properties of Darboux sums and integrals that

   $$U(f, P) - \int\limits_a^b f(x)\, dx < \dfrac{\varepsilon}{2} \text{ and } U(g, P) - \int\limits_a^b g(x) \, dx < \dfrac{\varepsilon}{2}.$$

   Therefore

   $$U(f, P) < \int\limits_a^b f(x)\, dx + \dfrac{\varepsilon}{2} \text{ and } U(g, P) < \int\limits_a^b g(x) \, dx + \dfrac{\varepsilon}{2}.$$

   The prior proof also shows that

   $$U(f + g, P) \leq U(f, P) + U(g, P).$$

   The properties of Darboux sums and integrals again show that

   $$\displaystyle \int\limits_a^b f(x) + g(x) \, dx \leq U(f + g, P).$$

   Combining this with the preceding inequalities gives us the following:

   $$\begin{array}{rcl} \displaystyle \int\limits_a^b f(x) + g(x) \, dx & \leq & U(f, P) + U(g, P) \\ \displaystyle \int\limits_a^b f(x) + g(x) \, dx & < & \displaystyle \int\limits_a^b f(x) \, dx + \int\limits_a^b g(x) \, dx + \varepsilon\end{array}$$

   Since $\varepsilon$ is arbitrary, we can conclude that

   $$\int\limits_a^b f(x) + g(x) \, dx \leq \int\limits_a^b f(x) \, dx + \int\limits_a^b g(x) \, dx.$$

   Part 2:

   We repeat this process with the lower Darboux sums:

   It follows from the properties of Darboux sums and integrals that

   $$\int\limits_a^b f(x)\, dx - L(f, P) < \dfrac{\varepsilon}{2} \text{ and } \int\limits_a^b g(x) - L(g, P) \, dx < \dfrac{\varepsilon}{2}.$$

   Therefore

   $$L(f, P) > \int\limits_a^b f(x)\, dx - \dfrac{\varepsilon}{2} \text{ and } L(g, P) > \int\limits_a^b g(x) \, dx - \dfrac{\varepsilon}{2}.$$

   The prior proof also shows that

   $$L(f + g, P) \geq L(f, P) + L(g, P).$$

   The properties of Darboux sums and integrals again show that

   $$\displaystyle \int\limits_a^b f(x) + g(x) \, dx \geq L(f + g, P).$$

   Combining this with the preceding inequalities gives us the following:

   $$\begin{array}{rcl} \displaystyle \int\limits_a^b f(x) + g(x) \, dx & \geq & L(f, P) + L(g, P) \\ \displaystyle \int\limits_a^b f(x) + g(x) \, dx & > & \displaystyle \int\limits_a^b f(x) \, dx + \int\limits_a^b g(x) \, dx - \varepsilon\end{array}$$

   Since $\varepsilon$ is arbitrary, we can conclude that

   $$\int\limits_a^b f(x) + g(x) \, dx \geq \int\limits_a^b f(x) \, dx + \int\limits_a^b g(x) \, dx.$$

   It follows by trichotomy that

   $$\int\limits_a^b f(x) + g(x) \, dx = \int\limits_a^b f(x) \, dx + \int\limits_a^b g(x) \, dx.$$

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7. Show that if $f, g \in \mathscr{R}([a, b]),$ then $f - g \in \mathscr{R}([a, b]).$

   Since the product of a real number and a Riemann integrable function is Riemann integrable, it follows that $-1g = -g \in \mathscr{R}([a, b]).$ Since the sum of Riemann integrable functions is Riemann integrable, it follows that $f - g \in \mathscr{R}([a, b]).$

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8. Show that if $f, g \in \mathscr{R}([a, b]),$ then $fg \in \mathscr{R}([a, b]).$

   Hint: Consider $f + g$ and composition by $h(x) = x^2.$

   Let $f, g \in \mathscr{R}([a, b]).$ It follows that

   $$f + g \in \mathscr{R}([a, b])$ \text{ and } f - g \in \mathscr{R}([a, b]).$$

   Let $h(x) = x^2.$ Because the the composition of a continuous function with a Riemann integrable function is Riemann integrable, it follows that

   $$h \circ (f+g) = (f + g)^2 \text{ and } h \circ (f - g) = (f - g)^2$$

   are also Riemann integrable. Because differences and scalar products of Riemann integrable functions are also Riemann integrable, we see that:

   $$\dfrac{1}{4}\left((f + g)^2 - (f - g)^2\right) \in \mathscr{R}([a, b])$$

   Expanding the algebra shows that

   $$\dfrac{1}{4}\left(f^2 + 2fg + g^2 - f^2 + 2fg - g^2\right) \in \mathscr{R}([a, b])$$

   which simplifies to

   $$fg \in \mathscr{R}([a, b]).$$

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9. Show that if $f, g \in \mathscr{R}([a, b])$ and $\inf(g([a, b])) > 0,$ then $\dfrac{f}{g} \in \mathscr{R}([a, b]).$

   Since $g \in \mathscr{R}([a, b]),$ it follows that $g$ is bounded. Combined with the fact that $\inf(g([a, b])) > 0,$ it follows that $\dfrac{1}{g([a, b])}$ is also bounded and nonzero. Since $h(x) = \frac{1}{x}$ is continuous on any domain without $0,$ the composition $h \circ g = \dfrac{1}{g}$ is Riemann integrable. Since the product of Riemann integrable functions is itself Riemann integrable, we conclude that $\frac{f}{g} \in \mathscr{R}([a, b]).$

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10. Give an example of functions $f$ and $g$ that are each Riemann integrable on $[a, b],$  $g > 0,$ but $\inf\limits_{x \in [a, b]}(g(x)) = 0$ such that $\frac{f}{g}$ is not Riemann integrable on $[a, b].$

    Consider the interval $[0, 4].$ Let

    $$f(x) = x - 2 \text{ and } g(x) = \left\{\begin{array}{ll} (x-2)^2 & \text{ where } 0 \leq x < 2\\ (x-2)^2 + 1 & \text{ where } 2 \leq x \leq 4 \end{array}\right..$$

    Then because $f$ is continuous and $g$ has a single discontinuity, both are Riemann integrable. However,

    $$\dfrac{f}{g}(x) = \left\{ \begin{array}{ll}\dfrac{1}{x-2} & \text{ where } 0 \leq x < 2\\ \dfrac{x - 2}{(x-2)^2 + 1} & \text{ where } 2 \leq x \leq 4 \end{array}\right.$$

    The quotient function $\frac{f}{g}$ is unbounded because it tends towards infinity as $x$ approaches $2$ from the left. Because $\frac{f}{g}$ is not bounded, it is not Riemann integrable.

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11. Subdivisibility of Domain: Show that if $f, \in \mathscr{R}([a, b])$ and $c \in (a, b),$ then

    $$\int\limits_a^b f(x) \, dx = \int\limits_a^c f(x) \, dx + \int\limits_c^b f(x) \, dx.$$

    First we show that $f \in \mathscr{R}([a, c])$ and $f \in \mathscr{R}([c, b]),$ then we show that the integral over $[a, b]$ is equal to the sum of the integrals over $[a, c]$ and $[c, b].$

    Part 1:

    Because $f \in \mathscr{R}([a, b]),$ for every $\varepsilon > 0$ there exists a finite interval subdivision $P_1$ such that $U(f, P_1) - L(f, P_1) < \varepsilon.$ Pick the partition $P_2 = \{[a, c], [c, b]\},$ and set $P = P_1 \# P_2.$ Then

    $$U(f, P) - L(f, P) < \varepsilon.$$

    Let

    $$P_a = \{ p \in P : p \cap [a, c] \neq \varnothing\} \text{ and } P_b = \{ p \in P : p \cap [c, b] \neq \varnothing\}.$$

    Then $P = P_a \cup P_b$ and

    $$U(f, P) = U(f, P_a) + U(f, P_b) \text{ and } L(f, P) = L(f, P_a) + L(f, P_b).$$

    Substituting these equalities into the inequality involving $\varepsilon$ shows us that

    $$U(f, P_a) - L(f, P_a) + U(f, P_b) - L(f, P_b) < \varepsilon.$$

    Since an upper Darboux sum is always greater than or equal to a lower Darboux sum of the same function and subdivision, we note that

    $$U(f, P_a) - L(f, P_a) \geq 0 \text{ and } U(f, P_b) - L(f, P_b) \geq 0.$$

    Therefore

    $$U(f, P_a) - L(f, P_a) < \varepsilon \text{ and } U(f, P_b) - L(f, P_b) < \varepsilon.$$

    Thus $f \in \mathscr{R}([a, c])$ and $f \in \mathscr{R}([c, b]).$

    Part 2:

    This portion of the proof is similar to the proof of the additivity property. We establish upper and lower bounds and show that they are the same to establish equality.

    Part 2.A:

    It follows from the properties of Darboux sums and integrals that

    $$U(f, P_a) - \int\limits_a^c f(x) \, dx < \varepsilon \text{ and } U(f, P_b) - \int\limits_c^b f(x) \, dx < \varepsilon.$$

    Therefore

    $$U(f, P_a) < \int\limits_a^c f(x) \, dx + \varepsilon \text{ and } U(f, P_b) < \int\limits_c^b f(x) \, dx + \varepsilon.$$

    Likewise, because

    $$U(f, P) = U(f, P_a) + U(f, P_b),$$

    the properties of Darboux sums and integrals show us that

    $$\int\limits_a^b f(x) \, dx \leq U(f, P_a) + U(f, P_b).$$

    Combining these facts shows us that

    $$\int\limits_a^b f(x) \, dx < \int\limits_a^c f(x) \, dx + \int\limits_c^b f(x) \, dx + 2\varepsilon.$$

    Since $\varepsilon$ is arbitrary, it follows that 

    $$\int\limits_a^b f(x) \, dx \leq \int\limits_a^c f(x) \, dx + \int\limits_c^b f(x) \, dx.$$

    Part 2.B:

    It follows from the properties of Darboux sums and integrals that

    $$L(f, P_a) - \int\limits_a^c f(x) \, dx > \varepsilon \text{ and } L(f, P_b) - \int\limits_c^b f(x) \, dx > \varepsilon.$$

    Therefore

    $$L(f, P_a) > \int\limits_a^c f(x) \, dx + \varepsilon \text{ and } L(f, P_b) > \int\limits_c^b f(x) \, dx + \varepsilon.$$

    Likewise, because

    $$L(f, P) = L(f, P_a) + L(f, P_b),$$

    the properties of Darboux sums and integrals show us that

    $$\int\limits_a^b f(x) \, dx \geq L(f, P_a) + L(f, P_b).$$

    Combining these facts shows us that

    $$\int\limits_a^b f(x) \, dx > \int\limits_a^c f(x) \, dx + \int\limits_c^b f(x) \, dx - 2\varepsilon.$$

    Since $\varepsilon$ is arbitrary, it follows that 

    $$\int\limits_a^b f(x) \, dx \geq \int\limits_a^c f(x) \, dx + \int\limits_c^b f(x) \, dx.$$

    It follows by trichotomy that

    $$\int\limits_a^b f(x) \, dx = \int\limits_a^c f(x) \, dx + \int\limits_c^b f(x) \, dx.$$

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12. Ordering Property: Show that if $f, g \in \mathscr{R}([a, b])$ and $f(x) \leq g(x)$ for all $x \in [a, b],$ then

    $$\int\limits_a^b f(x) \, dx \leq \int\limits_a^b g(x) \, dx.$$

    Let $P$ be a finite interval subdivision of $[a, b].$ Because $f(x) \leq g(x)$ for all $x \in [a, b],$ it follows from the definitions of infimum and supremum that

    $$L(f, P) \leq U(f, P) \leq L(g, P) \leq U(g, P).$$

    The properties of Darboux sums show that

    $$L(f, P) \leq \int\limits_a^b f(x) \, dx \leq U(f, P) \text{ and } L(g, P) \leq \int\limits_a^b g(x) \, dx \leq U(g, P).$$

    Combining the inequalities produces the desired result:

    $$\int\limits_a^b f(x) \, dx \leq \int\limits_a^b g(x) \, dx.$$

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