# Riemann Integration Part 1

In this post I’ll outline the basic theory of Riemann integration. Let ${f:[a,b]\rightarrow\mathbb R}$ be a bounded function, i.e. there exists ${M\in\mathbb R}$ such that ${|f(x)|\leqslant M}$ for all ${x\in[a,b]}$. A partition of ${[a,b]}$ is a finite subset ${\mathcal P = \{x_1, x_2, \ldots, x_m\}}$ such that ${a=x_1. The lower sum of a partition ${\mathcal P}$ is

$\displaystyle L_f(\mathcal P) := \sum_{i=1}^{m-1}\min_{x_i\leqslant x\leqslant x_{i+1}}f(x)(x_{i+1}-x_i).$

The upper sum of a partition ${\mathcal P}$ is

$\displaystyle U_f(\mathcal P) := \sum_{i=1}^{m-1}\max_{x_i\leqslant x\leqslant x_{i+1}}f(x)(x_{i+1}-x_i).$

Observe that for any partition ${\mathcal P}$, ${L_f(\mathcal P)\leqslant U_f(\mathcal P)}$, since ${\min_{x_i for each ${i}$. Given a partition ${\mathcal P}$, we say that ${\mathcal P'}$ is a partition of ${\mathcal P}$ if ${\mathcal P'\supset\mathcal P}$. Time for a theorem:

Theorem 1 If ${\mathcal P'}$ is a refinement of ${\mathcal P}$, then ${L_f(\mathcal P')\geqslant L_f(\mathcal P)}$ and ${U_f(\mathcal P')\leqslant U_f(\mathcal P).}$

Proof: Suppose ${\mathcal P = \{x_1, x_2, \ldots, x_m\}}$ and ${\mathcal P' = \mathcal P\cup\{y\}}$ where ${x_j for some ${1\leqslant j\leqslant m-1}$. Then

\displaystyle \begin{aligned} L_f(\mathcal P') - L_f(\mathcal P) &= \sum_{i=1}^{j-1} \min_{x_i

Since ${x_{j+1}-x_j = (x_{j+1}-y) - (y-x_j)}$ and

\displaystyle \begin{aligned} \min_{x_j

it follows that ${L_f(\mathcal P') - L_f(\mathcal P)\geqslant0}$, so that ${L_f(\mathcal P')\geqslant L_f(\mathcal P)}$. The proof that ${U_f(\mathcal P')\leqslant U_f(\mathcal P)}$ is entirely analogous. $\Box$

We’re now ready to define the integral. If for each ${\varepsilon>0}$ there exists a partition ${\mathcal P}$ such that ${U_f(\mathcal P)-L_f(\mathcal P)<\varepsilon}$, then ${f}$ is said to be Riemann integrable, and we write

$\displaystyle I(f) := \int_a^b f(x)\mathsf dx.$

For a basic example, constant functions are integrable. For if ${f(x) = c}$ for all ${x}$, then for any partition ${\mathcal P}$, ${U_f(\mathcal P)-L_f(\mathcal P)=0}$.

An equivalent, and easier to use, definition of Riemann integrability is this: ${f}$ is integrable iff there exists a sequence of partitions ${\{\mathcal P_n\}}$ with ${\mathcal P_n\subset\mathcal P_{n+1}}$ and

$\displaystyle \lim_{n\rightarrow\infty}U_f(\mathcal P)-L_f(\mathcal P)=0.$

In this case,

$\displaystyle \int_a^b f(x)\mathsf dx = \lim_{n\rightarrow\infty}U_f(\mathcal P_n) = L_f(\mathcal P_n).$

Another example: let ${f_n}$, ${n\geqslant1}$ be a sequence of functions on ${[0,1]}$ defined by ${f_1(x) = \chi_{[2^{-1},1]}}$ and ${f_{n+1}(x) = 2^{-n}\chi_{[2^{-(n+1)}, 2^{-n})}}$, where

$\displaystyle \chi_A(x) = \begin{cases} 1,& x\in A\\ 0,& x\notin A.\end{cases}$

Then

\displaystyle \begin{aligned} f&:=\lim_{n\rightarrow\infty} \sum_{k=1}^n f_k(x)\\ &= \chi_{[2^{-1},1]}+\sum_{n=1}^\infty 2^{-n}\chi_{[2^{-(n-1)}, 2^{-n})}\\ &= \begin{cases}1,& 2^{-1}\leqslant x\leqslant 1\\2^{-n},& 2^{-(n+1)}\leqslant x<2^{-n}\quad (n\geqslant1)\\ 0,& x=0 \end{cases}\end{aligned}

Clearly ${f}$ is bounded, as ${f(x)\leqslant 1}$ for all ${x}$. Define a sequence of partitions ${\{\mathcal P_n\}}$ by ${\mathcal P_1=\{0,1\}}$ and ${\mathcal P_{n+1}=\mathcal P_n\cup\{2^{-n} \}}$ for ${n\geqslant1}$. Then for ${n\geqslant1}$,

$\displaystyle L_f(\mathcal P_n)=\sum_{k=2}^n 2^{-k+1}2^{-k+2}=\sum_{k=2}^n2^{-2k+3},$

as ${L_f(\mathcal P_1)=0 = \sum_{k=2}^12^{-2k+3}}$, and for ${n\geqslant 1}$,

\displaystyle \begin{aligned} L_f(\mathcal P_{n+1}) &= L_f(\mathcal P_1) + \min_{2^{-(n+1)}\leqslant x<2^{-n}}f(x)(2^{-(n+1)}-2^{-n})\\ &= \sum_{k=2}^n 2^{-2k+3} + 2^{-n}2^{-(n+1)}\\ &= \sum_{k=2}^n 2^{-2k+3} + 2^{-2n+1}\\ &= \sum_{k=2}^{n+1} 2^{-2k+3} \end{aligned}

Evaluating this sum, we have

\displaystyle \begin{aligned} L_f(\mathcal P_n) &= \sum_{k=2}^{n} 2^{-2k+3}\\ &= \sum_{k=0}^{n-2} 2^{-2k-1}\\ & \frac12\sum_{k=0}^{n-2}\left(\frac14 \right)^k\\ &= \frac12\left(\frac{1 - \left(\frac14 \right)^{n-1}}{1-\frac14} \right)\\ &= \frac23\left(1-\left(\frac14\right)^{n-1} \right). \end{aligned}

It follows that

$\displaystyle \lim_{n\rightarrow\infty}L_f(\mathcal P_n) = \frac23.$

Since

$\displaystyle \min_{2^{-(n+1)}\leqslant x<2^{-n}}f(x) = 2^{-n} = \max_{2^{-(n+1)}\leqslant x<2^{-n}}f(x),$

${U_f(\mathcal P_n)=L_f(\mathcal P_n)}$ for all ${n}$. Hence ${f}$ is integrable, and

$\displaystyle \int_a^b f(x)\mathsf dx = \frac23.$

Now let’s compute the Lebesgue integral of this function. Since ${f_n\geqslant0}$ and ${f_n\leqslant f_{n+1}}$, by monotone convergence we have

\displaystyle \begin{aligned} \int_{[0,1]} f\mathsf dm &= \int_{[0,1]} \lim_{n\rightarrow\infty}\sum_{k=1}^n f_k\mathsf dm\\ &= \lim_{n\rightarrow\infty} \int_{[0,1]} \sum_{k=1}^n f_k\mathsf dm\\ &= \sum_{n=1}^\infty \int_{[0,1]} f_n\mathsf dm\\ &= \int_{[0,1]}\chi_{[2^{-1},1]} + \sum_{n=1}^\infty \int_{[0,1]} 2^{-n}\chi_{[2^{-(n+1)},2^{-n})}\mathsf dm\\ &= m([2^{-1},1]) +\sum_{n=1}^\infty 2^{-n}m([2^{-(n+1)},2^{-n}))\\ &= 1 - 2^{-1} +\sum_{n=1}^\infty 2^{-n}(2^{-n} - 2^{-(n+1)})\\ &= \frac12 + \sum_{n=1}^\infty 2^{-n}2^{-n-1}\\ &= \frac12 + \frac12\sum_{n=1}^\infty 2^{-2n}\\ &= \frac12\left(1 + \frac{\frac14}{1-\frac14} \right)\\ &= \frac12\left(1 +\frac13 \right)\\ &= \frac23. \end{aligned}