Real Analysis: Riemann Integration and Beyond


I blog for myself and hope that my public rantings will be useful to someone else.

They have!

In this entry, I will go over the material that my analysis exam will cover next week. This helps me to solidify what I already know.

To save time, I will not prove claims; for now, I will appeal to the proverbial it is left as an exercise to the reader.

This entry is a work in progress.

Riemann Integration


Let $f$ be a bounded function defined on an interval $\left[a,b\right]$. A partition of $\left[a,b\right]$ is a finite set $P=\left\{a=x_0<x_1<\cdots<x_n=b\right\}$. Set $\Delta_j=x_j-x_{j-1}$ and define the mesh of a partition $P$ as $\text{mesh}\left(P\right)=\max_{1\leqslant j\leqslant n}\Delta_j$. For each interval $\left[x_{j-1},x_j\right]$ of this partition, we define the maximum and minimum of $f$ on this interval by
$$M_j\left(f,P\right)=\sup_{x_{j-1}\leqslant x\leqslant x_j}f\left(x\right)\qquad\text{and}\qquad m_j\left(f,P\right)=\inf_{x_{j-1}\leqslant x\leqslant x_j}f\left(x\right).$$
Then we define the upper and lower sums of $f$ with respect to the partition $P$ by
$$U\left(f,P\right)=\sum_{j=1}^nM_j\left(f,P\right)\Delta_j\qquad\text{and}\qquad L\left(f,P\right)=\sum_{j=1}^nm_j\left(f,P\right)\Delta_j.$$
If, in addition, we are given a set of points $X=\left\{x_j':1\leqslant j\leqslant n\right\}$, where $x_j'\in\left[x_{j-1},x_j\right]$ for $1\leqslant j\leqslant n$, we define the Riemann sum
A partition $R$ is a refinement of a partition $P$ provided that $P\subseteq R$. If $P$ and $Q$ are two partitions, then $R$ is a common refinement of $P$ and $Q$ provided that $P\cup Q\subseteq R$.

Observe that, by definition, $L\left(f,P\right)\leqslant I\left(f,P,X\right)\leqslant U\left(f,P\right)$.

Define $L\left(f\right)=\sup_PL\left(f,P\right)$ and $U\left(f\right)=\inf_PU\left(f,P\right)$. A bounded function $f$ on a finite interval $\left[a,b\right]$ is called Riemann integrable if $L\left(f\right)=U\left(f\right)$. In this case, we write
for the common value.


Lemma. If $R$ is a refinement of $P$, then
$$L\left(f,P\right)\leqslant L\left(f,R\right)\leqslant U\left(f,R\right)\leqslant U\left(f,P\right).$$
Corollary. If $P$ and $Q$ are any two partitions of $\left[a,b\right]$, then
$$L\left(f,P\right)\leqslant U\left(f,Q\right).$$
In particular, we see that the set of numbers $\left\{L\left(f,P\right)\right\}$ is bounded above by any $U\left(f,Q\right)$. Hence by the completeness of $\mathbb R$, $\sup_PL\left(f,P\right)$ is defined. Moreover, $\sup_PL\left(f,P\right)\leqslant U\left(f,Q\right)$ for every partition $Q$. Therefore, $\inf_PU\left(f,P\right)$ is defined and $\sup_PL\left(f,P\right)\leqslant\inf_PU\left(f,P\right)$.

Riemann's Condition. Let $f$ be a bounded function on $\left[a,b\right]$. The following are equivalent:

1. $f$ is Riemann integrable.
2. For each $\varepsilon>0$, there is a partition $P$ so that $U\left(f,P\right)-L\left(f,P\right)<\varepsilon$.

Corollary. Let $f$ be a bounded real-valued function on $\left[a,b\right]$. If there is a sequence of partitions of $\left[a,b\right]$, $P_n$, so that
then $f$ is Riemann integrable. Moreover, if $X_n$ is any choice of points $x_{n,j}'$ selected from each interval of $P_n$, then
The evenly-spaced partition is often used. However, a strategically-chosen partition could be much better.

Theorem. Let $f$ be a bounded function on $\left[a,b\right]$. The following are equivalent:

1. $f$ is Riemann integrable.
2. For each $\varepsilon>0$, there is a partition $P$ so that $U\left(f,P\right)-L\left(f,P\right)<\varepsilon$.
3. For every $\varepsilon>0$, there is a $\delta>0$ so that every partition $Q$ such that $\text{mesh}\left(Q\right)<\delta$ satisfies $U\left(f,Q\right)-L\left(f,Q\right)<\varepsilon$.
4. For every $\varepsilon>0$, there is a $\delta>0$ so that every partition $Q$ such that $\text{mesh}\left(Q\right)<\delta$ and every choice of set $X=\left\{x_j':1\leqslant j\leqslant n\right\}$, where $x_j'\in\left[x_{j-1},x_j\right]$ satisfies
Theorem. Every monotone function on $\left[a,b\right]$ is Riemann integrable.

Theorem. Every continuous function on $\left[a,b\right]$ is Riemann integrable.

The Fundamental Theorem of Calculus


A function $f$ on $\left[a,b\right]$ has an antiderivative if there is a continuous function $F\left(x\right)$ on $\left[a,b\right]$ such that $F'\left(x\right)=f\left(x\right)$ for every point $x\in\left(a,b\right)$.


Theorem (Fundamental Theorem of Calculus). Let $f$ be a bounded Riemann integrable function on $\left[a,b\right]$, and let
$$F\left(x\right)=\int_a^xf\left(t\right)\,dt\qquad\text{for}\qquad a\leqslant x\leqslant b.$$
Then $F$ is a continuous function. If $f$ is continuous at a point $x_0$, then $F$ is differentiable at $x_0$ and $F'\left(x_0\right)=f\left(x_0\right)$.

Corollary. Let $f$ be a continuous function on $\left[a,b\right]$. Then $f$ has an antiderivative. Moreover, if $G$ is any antiderivative of $f$, then
Lemma. Suppose that $f$ is an integrable function on $\left[a,b\right]$ bounded by $M$. Then
$$\left|\int_a^bf\left(t\right)\,dt\right|\leqslant M\left(b-a\right).$$
Remark. A jump discontinuity in the integrand $f$ can result in a point where the integral is not differentiable. Nor is it the case that every differentiable function is an integral.

Normed Vector Spaces, Inner Product Spaces, and $L^p$ Norms


Let $V$ be a vector space over $\mathbb R$. A norm on $V$ is a function $\left\|\cdot\right\|$ on $V$ taking values in $\left[0,+\infty\right)$ with the following properties:

1. (positive definite) $\left\|x\right\|=0$ if and only if $x=0$,
2. (homogeneous) $\left\|\alpha x\right\|=\left|\alpha\right|\left\|x\right\|$ for all $x\in V$ and $\alpha\in\mathbb R$, and
3. (triangle inequality) $\left\|x+y\right\|\leqslant\left\|x\right\|+\left\|y\right\|$ for all $x,y\in V$.

We call the pair $\left(V,\left\|\cdot\right\|\right)$ a normed vector space.

Let $x=\left(x_1,x_2,\dots,x_n\right)\in\mathbb R^n$. Then
$$\left\|x\right\|_1=\sum_{i=1}^n\left|x_i\right|\qquad\qquad\qquad\qquad\left\|x\right\|_2=\left(\sum_{i=1}^n\left|x_i\right|^2\right)^{1/2}\qquad\qquad\qquad\qquad\left\|x\right\|_\infty=\max_{1\leqslant i\leqslant n}\left|x_i\right|$$
The above norms are well known.

For any normed vector space $\left(V,\left\|\cdot\right\|\right)$, the unit ball of $V$ is the set $\left\{x\in V:\left\|x\right\|\leqslant1\right\}$.

Let $K$ be a compact subset of $\mathbb R^n$, and let $C\left(K\right)$ denote the vector space of all continuous real-valued functions on $K$. The most natural and important norm on this vector space is the uniform norm given by
$$\left\|f\right\|_\infty=\sup_{x\in K}\left|f\left(x\right)\right|.$$
In a normed vector space $\left(V,\left\|\cdot\right\|\right)$, we say that a sequence $\left\{v_n\right\}_{n=1}^\infty$ converges to $v\in V$ if $\lim_{n\to\infty}\left\|v_n-v\right\|=0$. Equivalently, for every $\varepsilon>0$, there is an integer $N>0$ so that $\left\|v_n-v\right\|<\varepsilon$ for all $n\geqslant N$. This is written $\lim_{n\to\infty}v_n=v$.

Call $\left\{v_n\right\}_{n=1}^\infty$ a Cauchy sequence if for every $\varepsilon>0$, there is an integer $N>0$ so that $\left\|v_n-v_m\right\|<\varepsilon$ for all $n,m\geqslant N$.

Say that $\left(V,\left\|\cdot\right\|\right)$ is complete if every Cauchy sequence in $V$ converges to some vector $v\in V$. A complete normed space is called a Banach space.

For a normed vector space $\left(V,\left\|\cdot\right\|\right)$, we define the open ball with center $a\in V$ and radius $r>0$ to be $B_r\left(a\right)=\left\{v\in V:\left\|v-a\right\|<r\right\}$. A subset $U$ of $V$ is open if for every $a\in U$, there is some $r>0$ so that $B_r\left(a\right)\subseteq U$. A subset $C$ of $V$ is closed if it contains all of its limit points. That is, whenever $\left\{x_n\right\}_{n=1}^\infty$ is a convergent sequence of points in $C$ with limit $x=\lim_{n\to\infty}x_n$, then $x$ belongs to $C$.

A subset $K$ of a normed vector space $V$ is compact if every sequence $\left\{x_n\right\}_{n=1}^\infty$ of points in $K$ has a sub-sequence $\left\{x_{n_i}\right\}_{i=1}^\infty$ which converges to a point in $K$.

An inner product on a vector space $V$ is a function $\langle x,y\rangle$ on pairs $\left(x,y\right)$ of vectors in $V\times V$ taking values in $\mathbb R$ satisfying the following properties:

1. (positive definiteness) $\langle x,x\rangle\geqslant0$ for all $x\in V$ and $\langle x,x\rangle=0$ only if $x=0$.
2. (symmetry) $\langle x,y\rangle=\langle y,x\rangle$ for all $x,y\in V$.
3. (bilinearity) For all $x,y,z\in V$ and scalars $\alpha,\beta\in\mathbb R$, $\langle\alpha x+\beta y,z\rangle=\alpha\langle x,z\rangle+\beta\langle y,z\rangle$.

The $L^p$ norms on $C\left[a,b\right]$ for $1\leqslant p<\infty$ are defined by
For $1\leqslant p<\infty$, $\ell^p$ consists of the set of all finite sequences $a=\left\{a_n\right\}_{n=1}^\infty$ such that
Let $w\left(x\right)$ be a strictly positive piecewise continuous function on $\left[a,b\right]$. Then define a norm $\left\|\cdot\right\|_{L^p\left(w\right)}$ on $C\left[a,b\right]$ or even on the space $PC\left[a,b\right]$ of piecewise continuous functions by


Claim. A sequence $\left\{x_n\right\}_{n=1}^\infty$ in a normed vector space $V$ converges to a vector $x$ if and only if for each open set $U$ containing $x$, there is an integer $N$ so that $x_n\in U$ for all $n\geqslant N$.

Cauchy-Schwarz Inequality. For all vectors $x,y$ in an inner product space $V$,
$$\left|\langle x,y\rangle\right|\leqslant\left\|x\right\|\left\|y\right\|.$$
Equality holds if and only if $x$ and $y$ are collinear.

Corollary. For $f,g\in C\left[a,b\right]$, we have
Corollary. An inner product space $V$ satisfies the triangle inequality
$$\left\|x+y\right\|\leqslant\left\|x\right\|+\left\|y\right\|\qquad\text{for all}\qquad x,y\in V.$$
Moreover, if equality occurs, then $x$ and $y$ are collinear.

Corollary. Let $V$ be an inner product space with induced norm $\left\|\cdot\right\|$. Then the inner product is continuous (i.e., if $x_n$ converges to $x$ and $y_n$ converges to $y$, then $\langle x_n,y_n\rangle$ converges to $\langle x,y\rangle$).

Lemma. Let $A,B>0$. Then
$$A^tB^{1-t}\leqslant tA+\left(1-t\right)B\qquad\text{for all}\qquad0<t<1.$$
Moreover, equality holds for some (or all) $t$ only if $A=B$.

Hölder's Inequality. Let $w$ be a positive function on an interval $\left[a,b\right]$. Let $f\in L^p\left(w\right)$ and $g\in L^q\left(w\right)$ where $1<p<\infty$ and $1/p+1/q=1$. Then
Minkowski's Inequality. The triangle inequality holds for $L^p\left(w\right)$, that is,

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