Exam Practice Solutions

1. 1. The given collection contains $X$. It is closed under complements by direct verification: each of the sets $$\begin{array}{c}X\setminus \varnothing =XX\setminus A=X\setminus A\\ X\setminus (X\setminus A)=AX\setminus X=\varnothing \end{array}$$ belong to the collection. Lastly, given a sequence $E_1,E_2,\dots$ of sets in the collection their union is also in the collection. Indeed, we may assume all $E_n$ are non-empty. If some $E_n$ is $X$ then the union is $X$. We have reduced to the case where all $E_n$ are either $A$ or $X\setminus A$. The union of any such sequence is $A$ or $X\setminus A$ or $X$.
   2. Take $X=\mathbb{R}$ and $$\begin{array}{c}\mathcal{B}=\{\varnothing ,(-\mathrm{\infty },0),[0,\mathrm{\infty }),\mathbb{R}\}\\ \mathcal{C}=\{\varnothing ,(-\mathrm{\infty },4],(4,\mathrm{\infty }),\mathbb{R}\}\end{array}$$ both of which are σ-algebras by part (i). Their union $$\{\varnothing ,(-\mathrm{\infty },0),[0,\mathrm{\infty }),(-\mathrm{\infty },4],(4,\mathrm{\infty }),\mathbb{R}\}$$ is not a σ-algebra because the union $$(-\mathrm{\infty },0)\cup (4,\mathrm{\infty })$$ is not in the collection.

Comments

The first part is at a realtively low level of difficulty, requiring verification of the definition of a σ-algebra. The second part is at a medium level. Note that you do not have to have given a solution to the first part in order to use its result to answer the second part.

2. 1. Define $$g_n(x)=\{\begin{array}{ll}{\displaystyle \frac{1}{1+x^2}}& |x|\le n\\ 0& |x|>n\end{array}$$ for all $n\in \mathbb{N}$ and all $x\in \mathbb{R}$. Each $g_n$ is the product of the continuous function $f$ with the measurable function $1_{[-n,n]}$ so each $g_n$ is meaurable. The sequence $g_1(x),g_2(x),\dots$ is non-negative, non-decreasing and converges to $f(x)$ for all $x\in \mathbb{R}$. The monotone convergence theorem then gives $$\int f\mathsf{d}\lambda =\underset{n\to \mathrm{\infty }}{lim}\int g_n\mathsf{d}\lambda =\underset{n\to \mathrm{\infty }}{lim}\underset{-n}{\overset{n}{\int }}{\displaystyle \frac{1}{x^2+1}}\mathsf{d}x$$ because $g_n$ is continuous on $[-n,n]$ and zero outside $[-n,n]$ so the integral can be calculated using the Riemann integral. Now $$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\underset{-n}{\overset{n}{\int }}{\displaystyle \frac{1}{x^2+1}}\mathsf{d}x& =\underset{n\to \mathrm{\infty }}{lim}\arctan (x){|}_{-n}^n\\ & =\underset{n\to \mathrm{\infty }}{lim}2\arctan (n)=\pi \end{array}$$ so $$\int f\mathsf{d}\lambda =\pi$$ and $f$ does indeed belong to ${\mathsf{L}}^{\mathsf{1}}(\mathbb{R},\mathcal{B},\mu )$.
   2. Each $f_n$ is continuous and therefore measurable. Since $$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sin (\frac{x}{n})}{\frac{x}{n}}}=1$$ for all $x\ne 0$ we have $f_n\to f$ pointwise. We have $$|f_n(x)|=\left|{\displaystyle \frac{\sin (\frac{x}{n})}{\frac{x}{n}}}\right|{\displaystyle \frac{1}{1+x^2}}\le {\displaystyle \frac{1}{1+x^2}}=f(x)$$ for all $n\in \mathbb{N}$ and all $x\in \mathbb{R}$. From (i) we know that $f$ is integrable. We may therefore apply the dominated convergence theorem to get $$\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\lambda =\int \underset{n\to \mathrm{\infty }}{lim}{f}_n\mathsf{d}\lambda =\int f\mathsf{d}\lambda =\pi$$ as needed.

Comments

Both parts are at a medium level of difficulty. Take care to verify in your solutions that the hypotheses of any major theorems from the course are verified before applying them. Properties of elementary functions can be used without reference. Note that the second part tells you to use the dominated convergence theorem, so you must use it to get marks.

3. 1. $T$ is ergodic when every set $B\in \mathcal{B}$ satisfying $T^{-1}B=B$ has $\mu (B)\in \{0,1\}$.
   2. $T$ is ergodic if and only if every member $f$ of ${\mathsf{L}}^{\mathsf{1}}(X,\mathcal{B},\mu )$ satisfying $Tf=f$ is constant.
   3. If $T^2$ is not ergodic there is a non-constant function $g$ in ${\mathsf{L}}^{\mathsf{1}}(X,\mathcal{B},\mu )$ with $T^{2}g=g$ by (ii). Define $$f=g-Tg$$ and note that $f$ is non-zero, because if it were zero we would have $Tg=g$ and then $g$ would be constants. Now $$Tf=Tg-T^{2}g=Tg-g=-(g-Tg)=-f$$ so $f$ is an eigenfunction with eigenvalue $-1$.

Comments

The first two parts are recall from the notes. The third part is at a high-level of difficulty, being unlike previously seen problems and requiring an insight.

4. 1. If $T$ is ergodic then for every measurable and integrable function $f:X\to \mathbb{R}$ there is a set $\mathrm{\Omega }\subset X$ with $\mu (\mathrm{\Omega })=1$ such that $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))=\int f\mathsf{d}\mu$$ for all $x\in \mathrm{\Omega }$.
   2. We proved in class that $T$ is ergodic for $\mu$. Since $$f=1_{[1]}+1_{[01]}$$ it is simple and its integral is $\mu ([1])+\mu ([01])=\frac{3}{4}$. Thus $f$ is measurable and integrable. The pointwise ergodic theorem gives a set $\mathrm{\Omega }\subset \{0,1{\}}^{\mathbb{N}}$ with $\mu (\mathrm{\Omega })=1$ and $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))=\int f\mathsf{d}\mu ={\displaystyle \frac{3}{4}}$$ for all $x\in \mathrm{\Omega }$. Now $$\begin{array}{rl}& \sum _{n=0}^{N-1}f(T^n(x))\\ =& |\{1\le n\le N:x(n)=1\}|\\ & +|\{1\le n\le N:x(n)=0\text{ and }x(n+1)=1\}|\end{array}$$ so we deduce almost every infinitely repeated fair coin toss will have either a head or a tail followed by a head three-quarters of the time.

Comments

The first part is recall. The second part is at a low to medium level of difficulty. Take care when integrating $f$ as it is not given written as a simple function.