Converse of Rademacher's theorem. 
From Rademacher's theorem we know that 'every Lipschitz function ($\mathbb R^n \to \mathbb R$) is differentiable almost everywhere'.

My question is that "is the converse of the Rademacher's theorem true?".If not true then what will be a counterexample.
Please someone help..
Thank you..
 A: The converse states that an almost everywhere differentiable function is Lipschitz. 
This is not true: choose for example $f\colon x\mapsto \left\lVert x\right\rVert^2$. There are of course many other counter-examples. The idea is that we do not control the derivative and which may be unbounded.      
A: I will add my two cents on this subject for future readers since I find the content of this conversation quite misleading.
If you read the statement of Rademacher as you stated then of course everything breaks.
The problem is that the property of being a Lipschitz function is the existence of some $K>0$ such that
$$
\frac{|f(x)-f(y)|}{\|x-y\|}\leq K.
$$
This formula is basically saying that whenever the derivative exists it will be bounded by $K$.
So we can roughly identify the Lipschitz functions with functions with bounded derivatives, at least on a colloquial discourse.
Having pointed this out the statement of Rademacher theorem should include this little part and therefore stating that any Lipschitz function is differentiable almost everywhere and with derivative bounded by the Lipschitz constant of the function.
If you are familiar with Sobolev spaces, what this statement is saying is literally that we have an embedding
$$
\mathrm{Lip}(\mathbb R^d)\to W^{1,\infty}(\mathbb R^d),
$$
where if you aren't familiar with Sobolev spaces $W^{1,\infty}(\mathbb R^d)$ is the space of functions with bounded weak derivative, which could be seen as an almost everywhere standard derivative if you were to assume absolute continuity on all lines.
Now the converse statement to the Rademacher's theorem is when a function $f\in W^{1,\infty}(\mathbb R^d)$ is Lipschitz continuous.
That reduces to ask when it is continuous, since by having a continuous function with bounded almost everywhere derivativative we already have a Lipschitz function.
To solve this the Sobolev embeddings come to our aid by saying that if we have $d\leq pk$, $\alpha\in(0,1]$, and $r\in\mathbb N$ we have
$$
W^{k,p}(\mathbb R^d)\subseteq C^{r,\alpha}(\mathbb R^d)
$$
whenever
$$
\frac{1}{p}-\frac{k}{d}=-\frac{r+\alpha}{d}.
$$
By plugging $p=+\infty$ we obtain $r=0$ and $\alpha=1$ thus obtaining the desired identification.
