# Define the Riemann integral via trapezoids instead of rectangles

Let $I$ be an interval and $f\colon I \to \mathbb{R}$.

Recall that $f$ is called Riemann-integrable with integral $s$ if the following is true:

For all $\epsilon > 0$, there exists $\delta > 0$ such that for any tagged partition $x_0,\ldots,x_n$ of $I$ and $t_0,\ldots,t_{n-1}$ whose mesh is less than $\delta$, we have

$$\left|\sum_{i=0}^{n-1} f(t_i) (x_{i+1}-x_i) - s\right| < \epsilon$$

The intuitive idea which leads to the Riemann integral is that you approximate the "area under the curve" by rectangles. However one could also start with the idea to approximate it via trapezoids. So one could try to define the "trapezoid integral" via:

For all $\epsilon > 0$, there exists $\delta > 0$ such that for any partition $x_0,\ldots,x_n$ of $I$ whose mesh is less than $\delta$, we have

$$\left |\frac{1}{2} \sum_{k=0}^{n-1} \left( x_{k+1} - x_{k} \right) \left( f(x_{k+1}) + f(x_{k})\right) -s \right | < \epsilon$$

• Would this "trapezoid integral" be equivalent to the Riemann integral in the sense that a function is trapezoid integrable iff it is Riemann integrable and the integral s are equal in this case? If not, is one more general than the other?
• If not: Is it possible to make a slightly different definition of the integral starting from the trapezoid idea such that one can state such a theorem?
• Is it also possible to generalize the idea to a Newton-Cotes approach and also get a clear connection to the Riemann integral
• Is this type of "trapezoid" integral (or a generalization) known in the literature? If so, do you have a reference which states and proves theorems about the relation to the Riemann integral?

Note that I know the trapezoid Rule for approximating the Riemann integral but this is only for numerical approximations.

A rough argument, why Riemann integrable and trapezoidal integrable should be equivalent.

Here is the case of a monotone increasing integrand function $f$:

First is a left Riemann sum, then the trapezoidal sum, then the right Riemann sum.   (from Riemann sum)

We get $S_{L}(n) \le s \le S_{T}(n) \le S_{R}(n)$ which sandwiches the trapezoidal sum, so if the Riemann integral converges, both $S_L$ and $S_R$ converge towards $s$ and the trapezoidal sum will as well.

For the other direction one would need to find a finer right Riemann sum e.g. $$s \le S_R(n_2) \le S_T(n)$$ It should be possible to find such an $n_2$ because each trapezium (trapezoid) side between $(x_i, f(x_i))$ and $(x_{i+1},f(x_{i+1}))$ itself is a linear function which is Riemann integrabel. If one chooses the minimum of $n_2$ over all trapezoidal columns it should work.

• Thanks for the idea. How to get it for general functions? – Julia Mar 16 '15 at 15:13

One direction of the equivalence is easy.

Using the triangle inequality

$$\left|\frac1{2}\sum_{i=0}^{n-1}[f(x_i)+f(x_{i+1})](x_{i+1}-x_i)-s\right| \\ = \left|\frac1{2}\sum_{i=0}^{n-1}f(x_i)(x_{i+1}-x_i)-\frac1{2}I+\frac1{2}\sum_{i=0}^{n-1}f(x_{i+1})(x_{i+1}-x_i)-\frac1{2}I\right|\\\leqslant \frac1{2}\left|\sum_{i=0}^{n-1}f(x_i)(x_{i+1}-x_i)-s\right|+\frac1{2}\left|\sum_{i=0}^{n-1}f(x_{i+1})(x_{i+1}-x_i)-s\right|.$$

If $f$ is Riemann integrable, then given $\epsilon > 0$ there exists $\delta > 0$ such that for any partition $P$ with mesh less than $\delta$ each absolute value on the RHS is less than $\epsilon$.

Thus, $||P|| < \delta$ implies

$$\left|\frac1{2}\sum_{i=0}^{n-1}[f(x_i)+f(x_{i+1})](x_{i+1}-x_i)-s\right| < \epsilon.$$

Proving, for a bounded function, that integrability in the trapezoidal sense implies Riemann integrability is more difficult.

For a related question, see

The equivalence between Cauchy integral and Riemann integral for bounded functions

where reference is made to a 1915 paper by Gillespie -- proving indirectly that the existence of the Cauchy integral (a limit of left-hand sums) implies the existence of the Riemann integral. That proof is somewhat difficult to follow.

Direct Proof: Convergence of trapezoidal sums implies Riemann integrability

The trapezoidal sum is an average of left- and right-hand sums:

$$T(P,f) = \frac1{2}[S_L(P,f) + S_R(P,f)] = \frac1{2}\sum_{i=0}^{n-1}[f(x_i)+f(x_{i+1})](x_{i+1}-x_i).$$

Since the trapezoidal sums converge to $s$, for any $\epsilon > 0$ there is a partition $R = (x_0,x_1,\ldots,x_m)$ with $m$ subintervals such that

$$s - \epsilon/2 < T(R,f) < s + \epsilon/2.$$

Let $D = \sup_{x,y \in I}|f(x)-f(y)|$ be the maximum oscillation of $f$ on the interval $I$. Choose any partition $P = (y_0,y_1,\ldots,y_n)$ with $||P|| < \delta = \epsilon/(2Dm)$. Note that $m$ is the fixed number of subintervals in the partition $R$.

Taking the common refinement $Q = P \cup R$ we can show that an upper Darboux sum $U(P,f)$ satisfies

$$U(P,f) - S_L(Q,f) < m \delta D = \epsilon/2,\\U(P,f) - S_R(Q,f) < m \delta D = \epsilon/2 \\ \implies U(P,f) - T(Q,f) < \epsilon/2.$$

Since $Q$ is a refinement of $R$ we have $||Q|| \leqslant ||R||$ and $s - \epsilon/2 < T(Q,f) < s + \epsilon/2.$

Hence,

$$U(P,f) < T(Q,f) + \epsilon/2 < s + \epsilon.$$

By a similar argument we can show that for a lower Darboux sum $L(P,f)$,

$$L(P,f) > T(Q,f) - \epsilon/2 > s - \epsilon.$$

Therefore, for any partition $P$ with $||P|| < \delta$ and any corresponding Riemann sum $S(P,f)$ we have

$$s - \epsilon < L(P,f) \leqslant S(P,f) \leqslant U(P,f) < s + \epsilon \\ \implies |S(P,f) - s| < \epsilon.$$

• Thanks. Can you give more details for the nontrivial direction of the equivalence? – Julia Mar 17 '15 at 15:09
• @Julia: You're welcome. I added detail to the answer. Clearly Riemann integrability implies that trapezoidal sums converge to the integral since such a sum can be split into left and right-hand Riemann sums -- which for sufficiently fine mesh can be made arbitrarily close to the integral. – RRL Mar 17 '15 at 15:23
• By the "nontrivial direction" I mean the part that integrability in the trapezoidal sense implies Riemann integrability. – Julia Mar 17 '15 at 19:13
• Sorry, I misread. Yes - the question is whether convergence of the trapezoidal sum (average of right- and left-hand sums) as $||P|| \to 0$ implies Riemann integrability -- only knowing $f$ is bounded. Clearly for unbounded $f$, e.g. $f(x) = 1/\sqrt{x}$ then $f$ is not R-integrable yet a special sum -- the right-hand sum -- converges. The link I gave refers to Gillespie (1915) for a similar question pertaining to the Cauchy (left-hand) sum. You might try that approach- assume $f$ is not integrable, implying a set of discontinuities with non-zero content, and show a contradiction. – RRL Mar 18 '15 at 0:48
• Yes I am still interested. – Julia Mar 19 '15 at 12:04