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# Problems set III

Problem 12\_June\_2015\_1

Let $f$ and $f_{n}(n=1,2,...)$ be functions from $R \rightarrow R$ such that $f_{n}(x_{n})\rightarrow f(x)$ whenever $x_{n}\rightarrow x$. Show that $f$ is continuous.

Problem 12\_June\_2015\_2

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be monotonically increasing. Suppose $a, $a < f(a)$ and $f(b). Show that there exists $x\in (a,b)$ such that $f(x)=x.$

Problem 12\_June\_2015\_3

Let $\Omega =\mathbb{N},\mathcal{F}=$ power set of $\mathbb{N}$, $P\{n\}=2^{-n!}$ for $n=2,3,...$. [ $n!$ stands for the product of first $n$ positive integers]. Let $P\{1\}$ be determined by the condition $P(\Omega)=1$. If $X$ and $Y$ are independent random variables on $(\Omega ,\mathcal{F},P)$ show that at least one of them is almost surely constant.

## 6 thoughts on “Problems set III”

Solution for 2nd problem : Consider the non-empty set $A=\{x : f(x)\textgreater x,x\in[a,b]\}$. We claim that $m=\sup\{A\}$ is such that $f(m)=m.$ Suppose $f(m)=m+\epsilon\textgreater m.$ Clearly $m+\epsilon Let $\delta\textgreater 0$ be such that $a Since $m+\delta\notin A\Rightarrow f(m+\delta)\leq m+\delta$, which contradicts the increasing property since $f(m)=m+\epsilon \textgreater m+\delta.$ If $f(m)=m-\epsilon\textless m$, by the supremum property there exist $\delta\textgreater 0$ such that $m-\epsilon\textless m-\delta\in A\Rightarrow f(m-\delta)\textgreater m-\delta\textgreater m-\epsilon$ which is again a contradiction to the increasing property.Hence $f(m)=m$ and clearly $m\in(a,b)$

p.s There has been some writing problem in my earlier comment so i am reposting it. Also it would be good if we have a preview option(by installing plugin) so that we can see our comment before posting.

Consider the non-epmty set $A=\{x :$f(x)>x,x\in[a,b]\}\$. We claim that $m=sup\{A\}$ is such that $f(m)=m.$ Suppose $f(m)=m+\epsilon>m.$ Clearly $a0$ be such that
$a \leq m+m\delta+\delta$. If $f(m)=m-\epsilon0$ such that $m-\delta \geq m-\epsilon$ and $m-\delta\in A \Rightarrow f(m-\delta) \geq ;m-\delta$, which is again a contradiction since $f(m)=m-\epsilon \geq ;m-\delta$. Hence $f(m)=m$ and clearly $m\in [a,b]$