$$
\newcommand{\norm}[1]{\left\#1\right\}
\newcommand{\paren}[1]{\left(#1\right)}
\newcommand{\abs}[1]{\left\lvert#1\right\rvert}
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\ang}[1]{\left\langle#1\right\rangle}
\newcommand{\C}{\mathbb{C}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
$$
Due: January 19th, 2018
Math 142B Assignment 1

 Show that if $\mathcal{P_1}, \ldots, \mathcal{P_n}$ are partitions of the interval $[a, b]$, then there is a partition $\mathcal{P}$ of the interval $[a,b]$ which is a refinitement of every $\mathcal{P}_i$; that is, show any finite set of partitions has a common refinement.

Suppose $\paren{\mathcal{P_n}}_n$ is an infinite sequence of partitions of $[a, b]$.
Must there be a partition which is a refinement of every partition in the sequence?
(Make sure to check the definition of "partition" carefully.)

Let $f : [0, 2] \to \R$ be defined by $f(t) = t$.
Find a partition $\mathcal{P}$ of $[0, 2]$ such that
$$U(f, \mathcal{P})  L(f, \mathcal{P}) \lt \frac{1}{256}.$$


Suppose that $f : [a,b] \to \R$ is integrable, and $g : [a,b]\to\R$ is such that $g(x) = f(x)$ for all $x \in [a,b] \setminus\set{x_0}$.
Show that $g$ is integrable, and
$$\int_a^b f(t)\,dt = \int_a^b g(t)\,dt.$$

Using the previous part, show that if $f : [a,b] \to \R$ is integrable and $g : [a,b]\to\R$ is such that $g(x) = f(x)$ for all but finitely many points in $[a, b]$ then $g$ is integrable, and
$$\int_a^b f(t)\,dt = \int_a^b g(t)\,dt.$$
(Hint: use induction.)

Suppose that $f : [a, b] \to \R$ is integrable, and $g : [a,b] \to \R$ is such that $g(x) = f(x)$ except for at a sequence of points $(x_n)_n$.
Must $g$ be integrable?
Prove it, or provide a counterexample.

Suppose $f : [a,b] \to \R$ is bounded, $c \in (a, b)$, and $f$ is integrable on both $[a, c]$ and $[c, b]$.
Show that $f$ is integrable on $[a, b]$.


Find bounded functions $f, g : [0, 1] \to \R$ so that $f + g$ is integrable, but $f$ is not.

Find bounded functions $f, g : [0, 1] \to (0, \infty)$ so that $fg$ is integrable, but $f$ is not.

Bonus:
Let the function $f$ be defined as follows:
$$\begin{align*}
f : [0,1] &\longrightarrow \R\\
x &\longmapsto \begin{cases}\frac1q & \text{ if } x = \frac{p}{q} \in \Q \text{ with } p, q \in \N, \gcd(p,q) = 1{\color{red},} \text{ and } x \neq 0\\
0 & \text{ otherwise.}
\end{cases}
\end{align*}$$
Show that $f$ is integrable, and find its integral.
Notice that $\frac{d}{dt}\int_0^t f(x)\,dx \neq f(t)$.