Linear Algebra

Problem 1

Prove that if $T\in \mathcal{L}(V)$ is normal, then
$$\mbox{null}T^k=\mbox{null}T \mbox{ and } \mbox{range}T^k=\mbox{range}T$$

Solution:
easy to check that $\mbox{null}T\subset\mbox{null}T^k$
to prove an inclusion in the other direction, suppose that $v\in\mbox{null}T^k$
$T^kv=T(T^{k-1}v)=0\implies T^{k-1}v\in\mbox{null}T \mbox{ and } T^{k-1}v\in\mbox{range}T^{k-1}\subset\mbox{range}T$
but $V=\mbox{null}T\oplus(\mbox{null}T)^{\perp}=\mbox{null}T\oplus\mbox{range}T^*=\mbox{null}T\oplus\mbox{range}T$
the third equality holds because $T$ is normal
$T^{k-1}v\in \mbox{null}T\cap\mbox{range}T\implies T^{k-1}v=0$, inductively, we have $Tv=0$
This shows that $\mbox{null}T^{k}\subset\mbox{null}T$, completing the proof that $\mbox{null}T=\mbox{null}T^k$

Now we are going to show that $\mbox{range}T^k=\mbox{range}T$
$\mbox{range}T=\mbox{range}T^*=(\mbox{null}T)^{\perp}=(\mbox{null}T^k)^{\perp}=\mbox{range}(T^k)^*=\mbox{range}T^k$
first equality holds because T is normal, third equality is come from the previous proof, the last equality holds because $T^k$ is normal

Lebesgue Integral

Problem 1

Lebesgue Measure

Problem 1

Let $A_n$ be any sequence of subsets of $\mathbb{R}$
(i) if $A_n\uparrow A$ then $\lambda^*(A_n)\uparrow \lambda^*(A)$
(ii)$\lambda^*(\liminf A_n)\leq\liminf\limits_{n\to\infty} \lambda^*(A_n)$, where $\liminf A_n=\{x:x\in A_n \mbox{ultimately}\}$

solution:
(i)
Let  $F_n=\bigcup_{i=1}^{n}E_i$, $F=\cup E_n,A=\cup A_n$, where $E_n$ is measurable such that $A_n\subset E_n$, and $E_n-A_n$ is negligible (it is easy to prove such $E_n$ exists), clearly $A\subset F$,
$F_n\uparrow F$, therefore $\lambda^*(F_n)\uparrow \lambda^*(F)$
$\begin{align} \lambda^*(A_n)\leq\lambda^*(F_n)&\leq\lambda^*(A_n)+\lambda^*(F_n-A_n)\\ &\leq\lambda^*(A_n)+\sum_{i=1}^{n}\lambda^*(E_i-A_i)\\ &= \lambda^*(A_n) \end{align}$
$\lambda^*(F_n)=\lambda^*(A_n)\leq \lambda^*(A)\leq \lambda^*(F)$, and since $F_n\uparrow F$, we have $\lambda^*(A_n)\uparrow \lambda^*(A)$
(ii)
Let $B_n=\bigcap_{k\geq n}A_k$, then $B_n\uparrow B=\liminf A_n$
$\lambda^*(B_n)\leq \lambda^*(A_n)\Rightarrow \liminf\limits_{n\to\infty} \lambda^*(B_n)=\lambda^*(\liminf A_n)\leq \liminf\limits_{n\to\infty} \lambda^*(A_n)$
The equality follows from (i)

Problem 2

For $A\subset [a,b]\subset \mathbb{R}$, $\lambda^*(A)$ is also called the exterior measure of A in $[a,b]$ and is denoted $\lambda_e(A)$, the interior measure of A in $[a,b]$, denoted $\lambda_i(A)$, is defined by the formula
$$\lambda_i(A)=(b-a)-\lambda_e([a,b]-A)$$ in general, $\lambda_i(A)\leq\lambda_e(A)$ ; the set A is measurable if and only if $\lambda_i(A)=\lambda_e(A)$

solution:
$\Rightarrow$ is easy
$\Leftarrow$ We use $\lambda^*$ instead of $\lambda_e$ denote $I=[a,b]$, let $J$ be any bounded interval $$\begin{align*} \lambda^*(I)&=\lambda^*(I\cap A)+\lambda^*(I\cap A')\\ &=\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A\cap J')+\lambda^*(I\cap A'\cap J)+\lambda^*(I\cap A'\cap J') \\ &=\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A'\cap J)+\lambda^*(I\cap A\cap J')+\lambda^*(I\cap A'\cap J')\\ &\geq \lambda^*(I\cap J)+\lambda^*(I\cap J')\\ &=\lambda^*(I) \end{align*}$$
Therefore we have $\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A'\cap J)=\lambda^*(I\cap J)$, and it is easy to see that $\lambda^*(A\cap J)+\lambda^*(A'\cap J)=\lambda^*(J)$ Suppose $S$ is a subset of $\mathbb{R}$ for every $\varepsilon>0$, there exists $U,S\subset U=\bigcup J_n$ and $\sum \lambda^*(J_n)<\lambda^*(S)+\varepsilon$
$$\begin{align*} \lambda^*(A\cap S)+\lambda^*(A'\cap S)&\leq\lambda^*(A\cap U)+\lambda^*(A'\cap U)\\ &\leq \sum\lambda^*(A\cap J_n)+\sum\lambda^*(A'\cap J_n)\\ &=\sum(\lambda^*(A\cap J_n)+\lambda^*(A'\cap J_n))\\ &=\sum\lambda^*(J_n)\\ &<\lambda^*(S)+\varepsilon \end{align*}$$
Since $\varepsilon$ is arbitrary, we can conclude that A is measurable 

Problem 3

If A is a Lebesgue-measurable subset of $\mathbb{R}$ such that $\lambda(A)>0$, then the difference set $D=\{a-b:a,c\in A\}$ is a neighborhood of 0, that is, $(-\delta,\delta)\subset D$ for some $\delta>0$.