# Prove a Continuous Distribution Function is Uniformly Continuous

Let $F$ be the distribution function for a random variable $X$ and it is given that $F$ is continuous over the entire real line. Prove that $F$ is uniformly continuous over the real line.

My approach was something like this:

Take any two points $x$ and $y$ in $\mathbb R$. If $|x-y|>1$ then trivially $|F(x)-F(y)|\leq1<|x-y|$ hence $F$ is Lipschitz whenever $|x-y|>1$. So for all such points $x,y$ where $|x-y|>1$, $F$ is uniformly continuous.

However, I am stumped in the case where $|x-y|<1$. How do we finc a $\delta$ such that $|x-y|<\delta\implies|F(x)-F(y)|<\epsilon$, given any $\epsilon>0$?

• I think every bounded monotone continous functions are uniformly continous Jan 23, 2015 at 3:49

Given a bounded monotone function $$f:\mathbb{R}\to\mathbb{R}$$, $$\lim_{x\to-\infty} f(x)=a$$ and $$\lim_{x\to\infty} f(x)=b$$ exists, by monotone convergence theorem.

Given a continuous function $$f:\mathbb{R}\to\mathbb{R}$$ s.t. $$\lim_{x\to-\infty} f(x)=a$$ and $$\lim_{x\to\infty} f(x)=b$$ exists, then $$f$$ is uniformly continuous.

### Proof

$$\forall\epsilon>0$$:
$$\qquad$$ Let $$e=\epsilon/3$$.
$$\qquad$$ $$\exists M_1\in\mathbb{R}^+$$ s.t. $$f((-\infty,-M_1))\subseteq(a-e,a+e)$$.
$$\qquad$$ $$\exists M_2\in\mathbb{R}^+$$ s.t. $$f((M_2,\infty))\subseteq(b-e,b+e)$$.
$$\qquad$$ Let $$M=\max(M_1+1,M_2+1)$$.
$$\qquad$$ $$f$$ is continuous on $$[-M,M]$$, and so it is uniformly continuous.
$$\qquad$$ $$\exists d>0$$ s.t. $$\forall x,y\in [-M,M], |x-y|.
$$\qquad$$ $$\forall x,y\in\mathbb{R}$$:
$$\qquad\qquad$$ If $$|x-y|:
$$\qquad\qquad\qquad$$ If $$x,y\in [-M,M]$$, then $$|f(x)-f(y)|.
$$\qquad\qquad\qquad$$ If $$x,y\in (-\infty,-M)$$, then $$|f(x)-f(y)|\le|f(x)-a|+|a-f(y)|<2e$$.
$$\qquad\qquad\qquad$$ If $$x\in (-\infty,-M),y\in [-M,M]$$, then:
$$\qquad\qquad\qquad\qquad$$ $$|f(x)-f(y)|\le|f(x)-a|+|a-f(M)|+|f(M)-f(y)|<3e$$.
$$\qquad\qquad\qquad$$ All the cases are similar to the above, and so $$|f(x)-f(y)|<3e\le\epsilon$$.

• Can't we just use what you said earlier i.e. piecewise u.c. implies u.c.? Suppose I decompose $\mathbb R$ into lenghts of interval 1 each, then $F$ is u.c. on each such interval, then it is piecewise u.c. on all $\mathbb R$, so u.c. Won't this work? Jan 23, 2015 at 3:39
• @yedaynara No, there must be finitely many pieces for that one to work. Jan 23, 2015 at 3:40
• @yedaynara If not, any continuous function is uniformly continuous, but $\sin(x^2)$ is not. Jan 23, 2015 at 3:46
• Please explain the first sentence of ur answer. What do u mean by ends? Jan 23, 2015 at 3:47
• @yedaynara Alright. It's very precise now. Jan 23, 2015 at 3:53

$f$ is uniformly continuous on $\mathbb{R}\iff\forall\epsilon>0,\exists\delta>0$ s.t. $\forall x,y\in\mathbb{R},(|x-y|<\delta\implies|f(x)-f(y)|<\epsilon)$

If not, then:
$\qquad$ $\exists\epsilon>0$ s.t. $\forall\delta>0,\exists x,y\in\mathbb{R}$ s.t. $|x-y|<\delta$ and $|f(x)-f(y)|\ge\epsilon$.
$\qquad$ Substitute $\delta=1/n$ sequentially to obtain the sequences
$\qquad$ $(x_n),(y_n)$ which satisfy $|x_n-y_n|\to 0$ and $|f(x_n)-f(y_n)|\ge\epsilon$.
$\qquad$ If $(x_n)$ is not bounded above:
$\qquad\qquad$ Take a subsequence which $\to\infty$.
$\qquad\qquad$ The corresponding subsequence for $(y_n)$ also $\to\infty$.
$\qquad\qquad$ This contradicts the limit at infinity.
$\qquad$ So, $(x_n)$ is bounded above.
$\qquad$ Similarly, $(x_n)$ is bounded below.
$\qquad$ Take a convergent subsequence of $(x_n)$ which $\to t$.
$\qquad$ The corresponding subsequence for $(y_n)$ also $\to t$.
$\qquad$ This contradicts the continuity at $t$.

• Please see the solution I have posted. Jan 23, 2015 at 12:24

Not really an answer, but I can't comment.

You won't be able to prove that $F$ is Lipschitz continuous. Consider the Cantor function to see an example of a CDF which is not Lipschitz continuous. http://en.wikipedia.org/wiki/Cantor_function#Properties

Thanks to all who answered here. I thought of a solution, and whether it is correct or not, I leave it for you to judge.

$$F$$ is increasing, continuous, so consider any sequence $$\{x_n\}\subset[0,1]$$ such that $$x_n\to0$$.

Then, we will get a positive divergent sequence $$\{L_n\}$$ such that $$[x_n,1-x_n]\subset F([-L_n,L_n])$$. So $$F(L_n)\geq1-x_n$$ and $$F(-L_n)\leq x_n$$.

Thus, $$0\leq \lim_{n\to\infty} F(-L_n)\leq\lim_{n\to\infty} x_n=0$$ giving $$F(\lim_{n\to\infty}-L_n)=0$$ by Sandwich Theorem and continuity of $$F$$. Analogously it follows that $$F(\lim_{n\to\infty}L_n)=1$$.

Now we know that $$F$$ being continuous over $$\mathbb R$$, $$F$$ is uniformly continuous in the closed and bounded interval $$[-L_n,L_n]$$ for all $$n\in\mathbb N$$.

But, $$[-L_n,L_n]\subset[-L_{n+1},L_{n+1}],and\\\cup_{n\in \mathbb N}[-L_n,L_n]=\mathbb R.$$

$$F$$ is uniformly continuous in each $$[-L_n,L_n]$$ so $$F$$ is uniformly continuous in whole of $$\mathbb R$$

• The second half of your solution does not seem to use the first half at all, and I gave you a counterexample earlier to the logic in the second half, which was $\sin(x^2)$. It is uniformly continuous in $[-L,L]$ for all $L\in\mathbb{R}$ but not uniformly continuous on $\mathbb{R}$. You must use the existence of the limits to $\infty$ and $-\infty$ in a non-trivial way. Jan 23, 2015 at 16:58
• But $sinx^2$ is not monotone increasing, whereas the distribution function is. Jan 23, 2015 at 17:03
• Actually I am still having difficulty processing the answer you gave i.e. how to get the $\delta$ and all that. I would be so grateful if you could give a complete solution, please. Jan 23, 2015 at 17:07
• Yes, but then what about $x^3$? You need to use both the monotone property and the fact that it is bounded, for it to imply that the limits to $\infty$ and $-\infty$ to exist, and then you need to use this fact to prove that it is uniformly continuous. I'll update my answer soon... Jan 24, 2015 at 4:30
• Yes, I was wrong about the "mixing" of the deltas. In fact the delta is the exact same one you get from the uniform continuity of the interval (which you must choose large enough). Jan 24, 2015 at 5:51