A stationary distribution is a probability row vector $\pi$ satisfying $\pi P=\pi$. Its coordinates obey ${\pi }_{-}(1-p)={\pi }_{+}q$ and ${\pi }_{-}+{\pi }_{+}=1$. ThusBecause both transitions between distinct states have positive probability, the finite Markov chain is irreducible; its positive self-transition probabilities also make it aperiodic.

Condition on the intermediate state. Starting at $+1$, a return to $+1$ after two transitions occurs through $-1$ with probability $q(1-p)$ or through $+1$ with probability $(1-q{)}^2$. These alternatives are disjoint, so

By the addition rule for independent Poisson random variables, $X_n$ has the same distribution as a sum of $n$ independent variables $Z_j\sim \mathrm{Poisson}(\lambda )$. Each has expectation and variance $\lambda$. The central limit theorem therefore givesFor $\lambda =0$, every variable is zero, and the conclusion is the degenerate normal law with zero variance.

Independence and convolution giveby the binomial theorem. Thus

A Poisson distribution with parameter $\lambda \ge 0$ hasFor $\lambda =0$ it is concentrated at zero. For $\lambda >0$, shifting the index in the defining series gives $\mathbb{E}[X]=\lambda$ and $\mathbb{E}[X(X-1)]={\lambda }^2$, so $\mathrm{Var}X={\lambda }^2+\lambda -{\lambda }^2=\lambda$.

For the quotient map $q:X\to X/\sim$, the quotient topology is defined byTake the Hausdorff space $X=\mathbb{R}$ and identify two numbers precisely when they are both rational or both irrational. The quotient has two points. The inverse images of its singleton subsets are $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$, neither of which is open in $\mathbb{R}$, since each is dense and has dense complement. Hence the quotient has the indiscrete topology and is not Hausdorff.

The product topology on $X\times Y$ has as a basis of a topology the rectangles $U\times V$ with $U$ open in $X$ and $V$ open in $Y$; arbitrary unions of such rectangles are its open sets.

For distinct points $(x_1,y_1),(x_2,y_2)$, at least one coordinate differs. If $x_1\ne x_2$, disjoint open neighbourhoods $U_1,U_2$ in the Hausdorff space $X$ give disjoint open neighbourhoods $U_1\times Y,U_2\times Y$ of the two points. If the second coordinates differ instead, use $X\times V_1,X\times V_2$. Thus a product of two Hausdorff spaces is Hausdorff.

A sequence $x_n$ converges to $x$ if every open neighbourhood of $x$ contains all sufficiently late terms. A Hausdorff space separates any two distinct points by disjoint open neighbourhoods.

If a sequence had distinct limits $x,y$ in a Hausdorff space, choose disjoint open neighbourhoods $U$ of $x$ and $V$ of $y$. Convergence to each limit implies that all sufficiently late terms lie in both $U$ and $V$, contradicting $U\cap V=\varnothing$. Hence a convergent sequence in a Hausdorff space has a unique limit.

Let $d=h-g$. It vanishes outside the null set $E$, so part (e) proves $d$ is a measurable function, regardless of its values on $E$. Since the sum of two measurable functions is measurable, $h=g+d$ is measurable. This is the completeness of Lebesgue measure under changes on null sets.

If $a\ge 0$, the set $\{f>a\}$ is contained in $E$, since $f=0$ outside $E$. It is therefore a null set and is Lebesgue measurable. If $a<0$, its complement $\{f\le a\}$ is contained in $E$ and is measurable for the same reason. Complements of measurable sets are measurable. Thus every upper level set is measurable, proving $f$ is a measurable function.

A real measurable function is one whose inverse image of every open subset of $\mathbb{R}$ is Lebesgue measurable. Equivalently,These upper rays generate the Borel sigma-algebra, making this equivalent to requiring measurable inverse images of all Borel sets.

If $A\subseteq E$ and $E$ is a null set, every interval cover of $E$ also covers $A$. Thus monotonicity of Lebesgue outer measure gives $0\le m^{\ast }(A)\le m^{\ast }(E)=0$. Hence every subset of a null set is null, and in particular is Lebesgue measurable.