$f$ unbounded in $\mathbb{R}$ implies it cannot be in $L^2(\mathbb{R})$. Intuitively , I feel this must be true. I'm looking for a rigorous proof. So,I would like to confirm that there are no counter examples to begin with.
 A: That's wrong. Define
$$ f(x) = \begin{cases} x^{-1/3} & x \in (0,1)\\ 0 & \text{otherwise} \end{cases} $$
Then $f$ is unbounded, but 
$$ \int_0^1 f^2(x) \, dx = \int_0^1 x^{-2/3}\, dx = 3 $$

Addendum: 
To give another example, that shows that even if $f$ is bounded on compact sets, it can be unbounded, but still be $L^2$. Let 
$$ f = \sum_{n=1}^\infty n\cdot \chi_{[n,n+n^{-4})}$$
where $\chi_A$ denotes the characteristic function of a set $A$. Then $f$ is unbounded, but bounded on compact sets and 
$$ \int_{\mathbb R} f^2(x)\, dx = \sum_{n=1}^\infty \int_n^{n+n^{-4}} n^2 \,dx 
 = \sum_{n=1}^\infty \frac 1{n^2} < \infty $$ 
A: martini has answered your question quite thoroughly but just for fun note that $f$ could be unbounded in every open interval inside $(0,1)$ but yet still be in $L^{2}(\mathbb{R})$. Consider $f(x)=q$ if $x\in(0,1)$ and $x=\frac{p}{q}$, $p,q\in\mathbb{N}$, and $f(x)=0$ otherwise. Then $f=0$ almost everwhere and hence integrates to $0$. By shifting the previous function you can get a function that is unbounded in every open interval but integrates to $0$.
