Fifth assigment
Due Tuesday October 24, 2000 by 1PM (in 2-108)

All questions will be worth 5 points. With very brief answers!

1. Munkres, §13, no. 2 By Fubini's theorem the upper and lower integrals of $f(x,y)$ over $y\in B$ and are both integrable and have the same integral. Thus the difference is integrable over $A$ with zero integral. Since it is a non-negative function it must vanish except possibly on a set of measure zero. Thus the upper and lower integrals of $f$ over $y\in B$ must be equal off a set of measure zero.
2. Munkres, §13, no. 7 Serious hair-splitting is involved here.

   (a) Now, by definition, $E\subset \partial S$ (excuse my notation for the boundary). So if $x_0$ is in the interior of $S$ it is not in $D$ so $f_S=f$ nearby is continuous. If $x_0$ is in the interior of the complement then $f_S$ is locally zero, hence again continuous at $x_0.$ Thus we can suppose that $x_0\in\partial S.$ Since $x_0\notin D,$ $f$ is continuous as a function on $S$ nearby. On the otherhand since $x_0\notin E$ the limit at $x_0$ in $S$ is zero, hence the limit at $x_0$ along all sequences is $0,$ hence $f_S$ is continuous at $x_0.$

   (b) Here is the hair-splitting! If $f_S$ is continuous at $x_0$ the $x_0\notin D,$ since it is discontinuous even from $S$ there. On the other hand $x_0\notin E\setminus D$ since if $x_0\in E\setminus B$ then it is a non-isolated boundary point at which the limit of $f$ from $S$ either fails to exist of is non-zero; either way $f_S$ cannot be continuous there.

   (c) Each point $p\in B$ is a point in $S$ which is isolated, so is at a positive distance from $S\setminus\{p\}.$ Hence $B$ is discrete in the sense that each point of $B$ is positive distance from the rest of $B.$ Since $B$ is bounded, such a set is countable, there being only a finite number of points distant at least $1/n$ from the others, for each $n.$

   (d) So if $D$ and $E$ have measure zero, then $f_S$ is integrable by (a). If $f$ is integrable on $S,$ i.e. $f_S$ is integrable, then by (b) $D\cup(E\setminus B)$ has measure zero. Thus $D$ and $E\setminus B$ have measure zero, but by (c) $B$ is countable, so this is equivalent to the condition that both $D$ and $E$ have measure zero.

3. Spivak, no. 3-34. Let $g_1,g_2:\mathbb{R}^2\longrightarrow \mathbb{R}$ be continuously differentiable and suppose $D_1g_2=D_2g_1.$ Let

   $$f(x,y)=\int_0^xg_1(t,0)dt+\int_0^yg_2(x,t)dt.$$

   
   
   Show that $D_1f(x,y)=g_1(x,y).$ By the Fundamental Theorem of Calculus,

   $$D_1f(x,y)=g_1(x,0)+\int_0^yD_1g_2(x,t)dt= g_1(x,0)+\int_0^yD_2g_1(x,t)dt=g_1(x,0)+\big[g_2(x,t)\big]_0^y=g_2(x,y).$$

   
   

4. Spivak, no. 3-37. (a) Suppose that $f:(0,1)\longrightarrow \mathbb{R}$ is a non-negative continuous function. Show that $\int_0^1f$ exists if and only if $\lim_{\epsilon \to0}\int_\epsilon ^{1-\epsilon }f$ exists.

   (b)

5. Spivak, no. 3-38.