# Proof Verification: Comparison Property for Integrals

Theorem: Let $f,g:[a,b] \rightarrow \mathbb{R}$ be two (Darboux) integrable functions such that $f(x)>g(x)$ for all $x \in (a,b)$. Then $$\int_a^b f(x) \,dx>\int_a^b g(x) \, dx$$

I would like to note I'm working with the definition of the integral introduced by Thomas Apostol in his Calculus text. This I believe corresponds to the definition of the Darboux integral. The author, moreover, describes the proof as "not easy" (page 81), but I think it is. Why doesn't the following work?

We note that the above theorem is clear for step functions defined (on the same partition) on some interval. On each subinterval, the "rectangle" for $f$ will invariably have greater area than the "rectangle" for $g$. Hence the sum of the area of the rectangles will also be greater.

We partition $[a,b]$ into ${x_0, x_1...x_n}$. Define step functions $h,j,m$ satisfying, on each subinterval, $f(x)>m(x)>h(x)>j(x)>g(x)$ within the subintervals.

Note that it suffices to show $\sup\{\int_{a}^{b} s(x) \ dx | s \leq g\} = \underline{\int_{a}^{b}} g(x) \ dx < \overline{\int_{a}^{b}} f(x) \ dx = \inf\{\int_{a}^{b} t(x) \ dx | t \geq f\}$ where $s$ and $t$ are step functions satisfying $s(x) \leq g(x)$ and $t(x) \geq f(x)$.

We have inequalities $h(x)>g(x)\geq s(x)$. By our preliminary remark, $\int_{a}^{b} h(x) \ dx>\int_{a}^{b} s(x) \ dx$ for $s$ satisfying $s(x) \leq g(x)$. This implies the integral of $h$ is an upper bound for $\{\int_{a}^{b} s(x) \ dx | s \leq g\}$ but not the supremum since the integral of $j$ is closer. We thus have the inequality $\int_{a}^{b} h(x) \ dx> \underline{\int_{a}^{b}} g(x) \ dx$

Thus, now we merely must show $\inf\{\int_{a}^{b} t(x) \ dx | t \geq f\} = \overline{\int_{a}^{b}} f(x) \ dx \geq \int_{a}^{b} h(x) \ dx$. By our preliminary remark, $\int_{a}^{b} h(x) \ dx<\int_{a}^{b} t(x) \ dx$ for $t$ satisfying $t(x) \geq f(x)$. So we have the integral of $h$ as a lower bound for $\{\int_{a}^{b} t(x) \ dx | t \geq f\}$ but not the infimum since the integral of $m$ is closer. This, in fact, proves the stronger inequality $\overline{\int_{a}^{b}} f(x) \ dx > \int_{a}^{b} h(x) \ dx$.

• You don't say how $x_0, \dots, x_n$ are chosen. You might very well have $n = 1$ with this way of introducing them. – David Feb 20 '16 at 5:37

## 2 Answers

The proof you gave doesn't make sense to me. It's not clear at all how you define the points $x_i$, or how you know they can be chosen in such a way that $m, h, j$ exist. Below I discuss two proofs of the result you've stated, of which only the second is self-contained.

First note that by considering $f(x) - g(x)$ if necessary, you may assume without loss of generality that $g(x) = 0$. Assume therefore that $f$ is Riemann-integrable on $[a,b]$ and that $f(x) > 0$ there. The result then follows directly from: the Lebesgue criterion for Riemann-integrability (which came up in a previous question of yours), which guarantees that a Riemann-integrable function is continuous at "most" points in a certain sense, hence at at least one point $c$; and Ex. 7 of 3.20, which shows that if $f(x) \geq 0$ and $f$ is continuous at some point $c$ where $f(c) > 0$, then $\int_a^b f > 0$. (For a self-contained proof along these lines, see Exercises 30, 31 of Chapter 13 in Spivak's Calculus, 3rd edition.)

However, you will probably be interested in having a proof that doesn't rely on the Lebesgue criterion, so here is one.

Assume, for a contradiction, that $f(x) > 0$ on $[a,b]$ and $\int_a^b f = 0$. We must have $\int_c^d f = 0$ for any subinterval $[c,d]$ of $[a,b]$, since $\int_a^b f = \int_a^c f + \int_c^d f + \int_d^b f$.

Because the integral of $f$ is zero, there exists some step function $\varphi$ defined on $[a,b]$ such that $f(x) \leq \varphi(x)$ on $[a,b]$ and $\int_a^b \varphi < \varepsilon = b - a$. There must be some nontrivial subinterval $[a_1,b_1]$ on which we have $\varphi(x) \leq 1$. Consequently, $f(x) \leq 1$ on this interval.

Assume by induction that we have constructed sequences $a \leq a_1 \leq a_2 \leq \dots \leq a_{n-1} < b_{n-1} \leq b_{n-2} \leq \dots \leq b_1 \leq b$ such that for $k = 1, 2, \dots, n-1$, we have $f(x) \leq \frac{1}{k}$ on $[a_k,b_k]$. We will extend the sequence to include new terms $a_n$ and $b_n$.

Since $\int_{a_{n-1}}^{b_{n-1}} f = 0$, there is some step function $\varphi$ defined on $[a_{n-1},b_{n-1}]$ satisfying $f \leq \varphi$ on $[a_{n-1},b_{n-1}]$ and $\int_{a_{n-1}}^{b_{n-1}} \varphi < \varepsilon = \frac{1}{n}(b_{n-1} - a_{n-1})$. Thus there is some nontrivial subinterval $[a_n,b_n]$ of $[a_{n-1},b_{n-1}]$ on which $f(x) \leq \varphi(x) \leq \frac{1}{n}$, extending the construction to $k = n$ and hence, by induction, indefinitely.

Now let $r = \sup \{a_n \colon n \geq 1\}$. For every $n$, we have $r \in [a_n,b_n]$, so $f(r) \leq \frac{1}{n}$. Therefore $f(r) \leq 0$, a contradiction.

Note. The book assumes $f(x) > 0$ on $[a,b]$. Your statement assumes this only on $(a,b)$. My proof is valid in the first case. For your statement, apply the first case to some subinterval $[c,d]$ of $(a,b)$.

We partition $[a,b]$ into $x_0,x_1\cdots x_n$. Define step functions $h,j,m$ satisfying, on each subinterval, $f(x)>m(x)>h(x)>j(x)>g(x)$ within the subintervals.

Are you sure such functions exists for each subdivision?

• WLOG, suppose $f$ and $g$ are monotone. If increasing, we specify the partition s.t for all $x_i$, $f(x_i)>g(x_{i+1})$; given that the functions are continuous at the interval points and some neighbourhood containing them, we can make such a partition. We can do the analogous thing if they're both decreasing. Then we can "squeeze" the step functions between those values. If one is decreasing and the other is increasing, there shouldn't be any issues, since the minimum of $f$ is precisely at the maximum of $g$ (within an interval) and thus greater than it. – MathematicsStudent1122 Feb 19 '16 at 9:32
• The only issues are the endpoints, at which they can take the same value, so $f(x_i)>g(x_{i+1})$ may not hold. To deal with this, we can take the integral not from $a$ to $b$ but $a+ \epsilon$ to $b - \epsilon$ noting that, for example, $\int_{a}^{a+\epsilon} f(x) - g(x) \ dx < \int_{a}^{a+\epsilon} \max_{a \leq x \leq a+\epsilon}({f(x) - g(x)}) \ dx$ And $\int_{a}^{a+\epsilon} \max_{a \leq x \leq a+\epsilon}{f(x) - g(x)} \ dx = \epsilon(\max_{a \leq x \leq a+\epsilon}{f(x) - g(x)}) < \delta$ for all $\delta>0$ and $\epsilon$ sufficiently small. – MathematicsStudent1122 Feb 19 '16 at 9:33
• The case with $b$ is similar. Consequently, if the statement holds for $(a+\epsilon, b-\epsilon)$ for all $\epsilon$, it holds for $(a,b)$. – MathematicsStudent1122 Feb 19 '16 at 9:33
• note the second inequality in the second comment should have a $\leq$ rather than a $<$. – MathematicsStudent1122 Feb 19 '16 at 9:44
• What do you mean, "without loss of generality, suppose $f$ and $g$ are both monotone"? How do you know there are any intervals at all on which $f$ is monotone? – David Feb 20 '16 at 5:35