$$ \newcommand{\norm}[1]{\left\|#1\right\|} \newcommand{\paren}[1]{\left(#1\right)} \newcommand{\abs}[1]{\left\lvert#1\right\rvert} \newcommand{\set}[1]{\left\{#1\right\}} \newcommand{\ang}[1]{\left\langle#1\right\rangle} \newcommand{\C}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\N}{\mathbb{N}} $$
Due: January 29th, 2018

Math 142B Assignment 2

    1. Suppose that $f : [a,b]\to\R$ is integrable. Show that $|f|$ is integrable. (Hint: notice that for any $x, y \in [a,b]$, $|f(x)| - |f(y)| \leq |f(x)-f(y)|$.)
    2. Suppose that $f, g : [a,b]\to\R$ are integrable. Show that $\min(f, g)$ and $\max(f,g)$ are integrable. (Hint: there is a reason that this is Part B.)
  1. Suppose $f : [a,b]\to[0,\infty)$ is continuous. Show that $\int_a^b f = 0$ if and only if $f(t) = 0$ for all $t \in [a,b]$.
  2. Suppose $f$ is defined as follows: $$\begin{align}f : [4, 12] &\longrightarrow \R\\ t &\longmapsto \begin{cases} 3 & \text{ if } 5 \lt t \leq 10 \\ 1 & \text{ otherwise.}\end{cases}\end{align}$$ Show that $f$ does not have an antiderivative.

    (Hint: There are several ways of solving this problem. One may take a direct approach: first, show that if $f$ does have an antiderivative, then it must be a piecewise linear function with different slopes on the different pieces; but there is no way such a function could be differentiable where the slope changes. In fact, it is possible to prove a much more general statement: any function with a jump discontinuity has no antiderivative.)

  3. Suppose that $f : \R \to \R$ has a continuous second derivative.
    1. Show that $f(t) = f(0) + f'(0)t + \int_0^t (t-x)f''(x)\,dx$ for all $t$. (Hint: compare the second derivatives of both sides.)
    2. Show that if $G : \R \to \R$ is defined by $G(t) = \int_0^t(t-x)f(x)\,dx$, then $G''(t) = f(t)$ for all $t \in \R$.
  4. Suppose that $f : [a, b]\to\R$ is continuous, and that $F : [a,b] \to \R$ is continuous, differentiable on $(a,b)$, and $F'(z) = f(z)$ for $z \in (a,b)$. Use the second fundamental theorem to prove the first fundamental theorem by showing that $$\frac{d}{dt}\paren{F(t) - \int_a^t f(x)\,dx} = 0.$$

    Clarification:

    First, show the identity above. Then, use this to prove the first fundamental theorem: if $F : [a, b] \to \R$ is continuous, differentiable on $(a, b)$, and $F'$ is continuous and bounded on $(a, b)$ then $$\int_a^b F'(x)\,dx = F(b) - F(a).$$

  5. Bonus:

    A set $S \subset \R$ is said to be null if for any $\epsilon \gt 0$ there is a sequence of open intervales $\paren{(a_n, b_n)}_n$ such that their union contains $S$ and their combined length is less than $\epsilon$: $$S\subseteq \bigcup_{n=1}^\infty (a_n, b_n) \qquad\qquad\text{ and }\qquad\qquad \sum_{n=1}^\infty b_n-a_n \lt \epsilon.$$

    1. Suppose that $f : [a, b] \to \R$ is bounded, and that $\set{x \in [a,b] : f \text{ is discontinuous at } x}$ is null. Show that $f$ is integrable.
    2. Show that if $S \subset \R$ is countable, then $S$ is null.

    As it turns out, there are null sets which are not countable; the Cantor set is a famous example. Somewhat more surprisingly, the function which is $1$ on the Cantor set and $0$ on its complement is discontinuous precisely on the Cantor set, and therefore integrable by this problem. On the last assignment's bonus problem we showed that there was an integrable function with a countable set of discontinuities; now there is even one with an uncountable set of discontinuities.