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<x_2<\cdots<x_m=b}. 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<x<x_{i+1}}f(x)\leqslant\max_{x_i<x<x_{i+1}}f(x)} 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<y<x_{j+1}} 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<x<x_i+1} f(x)(x_{i+1}-x_i) + \min_{x_j<x<y}f(x)(y-x_j) + \min_{y<x<x_{j+1}}f(x)(x_{j+1}-y)\\ &\quad - \sum_{i=1}^{m-1}\min_{x_i\leqslant x\leqslant x_{i+1}}f(x)(x_{i+1}-x_i)\\ &= \min_{x_j<x<y}f(x)(y-x_j) + \min_{y<x<x_{j+1}}f(x)(x_{j+1}-y) - \min_{x_j<x<x_{j+1}}f(x)(x_{j+1}-x_j). \end{aligned}

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

\displaystyle  \begin{aligned} \min_{x_j<x<y}f(x) &\geqslant \min_{x_j<x<x_{j+1}}f(x)\\ \min_{y<x<x_{j+1}}f(x) &\geqslant \min_{x_j<x<x_{j+1}}f(x),\\ \end{aligned}

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}

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