Using Squeeze Theorem for Riemann Integral

Question

Consider the function $f : [1, 3] → \mathbb{R}$ defined by $f(x) = 2x + 1, ∀x ∈ [1, 3]$. Use the Squeeze Theorem for Riemann integral to prove that $f ∈ \mathcal R[1, 3]$, the set of Riemann integrable functions on $[1,3]$.

I tried to look up examples for squeeze theorem but I can't find any. I understand the definition but I am lost on how to apply them.

Answer 1

Let's show $f$ is Riemann integrable on $[1,3]$. To do this, we will write the integral as the limit of a Riemann sum and show it converges using the Squeeze Theorem. To construct the Riemann sum, first note that we are dealing with rectangles having width $$\Delta x=\frac{3-1}{n}=\frac{2}{n}. $$ Then the $x$ value of the $i^\text{th}$ rectangle is given by $$x_i=1+i\cdot\Delta x=1+\frac{2i}{n}.$$ We also need the value of $f(x_i)$, which is given by $$f(x_i)=2(x_i)+1=2\left(1+\frac{2i}{n}\right)+1=3+\frac{4i}{n}.$$ Putting this all together, we have (by definition of the definite Riemann integral), $$ \int_1^3f(x)dx=\int_1^3(2x+1)dx=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x=\lim_{n\to\infty}\sum_{i=1}^n\left(3+\frac{4i}{n}\right)\frac{2}{n}. $$ It remains to use the Squeeze Theorem to show the limit $$ \lim_{n\to\infty}\sum_{i=1}^n\left(3+\frac{4i}{n}\right)\frac{2}{n} $$ converges. To simplify things, note that $$ \sum_{i=1}^n\left(3+\frac{4i}{n}\right)\frac{2}{n}=\frac{2}{n}\left(\sum_{i=1}^n3+\frac{4}{n}\sum_{i=1}^ni\right)=\frac{2}{n}\left(3n+\frac{4}{n}\frac{n(n+1)}{2}\right)=\frac{4}{n}+10. $$ So we need to find two convergent sequences $\{a_n\}$ and $\{b_n\}$ such that $$ \lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n $$ and $$ a_n\leq \frac{4}{n}+10\leq b_n $$ for all $n\in\mathbb{N}$. We can simply take $$ a_n=\frac{3}{n}+10\quad\text{and}\quad b_n=\frac{5}{n}+10. $$ Can you conclude from here using the Squeeze Theorem?

Answer 2

Since $f$ is an increasing function, given a partition $P=\{1=x_0<x_1<\ldots<x_n=3\}$, the lower and upper Riemann sums, $L(f,P)$ and $U(f,P)$, are easy to calculate:

\begin{align*} L(f,P) &= \sum_{i=1}^n f(x_{i-1})(x_i-x_{i-1}) \\ &= \sum_{i=1}^n (2x_{i-1}+1)(x_i-x_{i-1}) \\ &= 2\sum_{i=1}^nx_{i-1}(x_i-x_{i-1}) + \sum_{i=1}^n(x_i-x_{i-1}) \\ &= 2\sum_{i=1}^nx_{i-1}(x_i-x_{i-1}) + x_n - x_0 \\ &= 2\sum_{i=1}^nx_{i-1}(x_i-x_{i-1}) + 2, \end{align*}

and similarly

$$ U(f,P) = 2\sum_{i=1}^nx_i(x_i-x_{i-1}) + 2. $$

Since $x_{i-1} < \frac{x_{i-1}+x_i}2 < x_i$, one has $L(f,P) < I < U(f,P)$, where

$$ I:=\sum_{i=1}^n(x_{i-1}+x_i)(x_i-x_{i-1}) + 2 = \sum_{i=1}^n(x_i^2-x_{i-1}^2) + 2 = x_n^2 - x_0^2 + 2 = 10.$$

In fact, we can calculate $I - L(f,P)$ and $U(f,P) - I$ explicitly:

$$ I - L(f,P) = 2\sum_{i=1}^n\Big[\tfrac{x_{i-1}+x_i}2 - x_{i-1}\Big](x_i-x_{i-1}) = \sum_{i=1}^n(x_i-x_{i-1})^2 $$

and similarly

$$ U(f,P) - I = \sum_{i=1}^n(x_i-x_{i-1})^2. $$

Since $\sum_{i=1}^n(x_i-x_{i-1})^2 \le \|P\|\sum_{i=1}^n(x_i-x_{i-1}) = 2\|P\|$, where $\|P\|:=\max_{1\le i\le n}(x_i-x_{i-1})$ is the mesh of the partition, we have

$$I - 2\|P\| \le L(f,P) < I < U(f,P) \le I + 2\|P\|.$$

Letting $\|P\|\to0$ and applying the squeeze theorem, one sees that $f\in\mathcal R[1,3]$ with $\int_1^3f(x)dx = I = 10$.