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consider $\Omega=(-1,1)$ and let $f$ be the Heaviside function

$$f(x) = \begin{cases} 1,&\text{if } 0 \leq x < 1\\ 0,&\text{if } -1 < x < 0.\end{cases}$$

Then $f'(x)$ defined by (in distribution sense)

$$\langle f',v\rangle :=-\int_{-1}^1 f(x)v'(x)\,dx$$

In fact, $f'$ is the Delta function, since

$$\langle f',v\rangle = -\int_0^1 v'(x) \, dx = v(0)-v(1) = v(0).$$ Now how to prove that $f'$ is not an element of $L^2(\Omega)$?

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    You can prove that $f'$ is supported only in $0$. Since that is a set of measure $0$ it would be the zerofunction if it was a function at all. – 2017-01-19

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Suppose $f'\in L^2$. Then $|\int_{-1}^1 f'v| \le C\|v\|_2$ where $C=\|f'\|_2$. On the other hand, this integral is $-v(0)$, so
$$ |v(0)| \le C\|v\|_2$$ But this is blatantly false, since a test function with $v(0)=1$ can have arbitrarily small $L^2$ norm. E.g., let $v$ be any test function with $v(0)=1$ and shrink it as $v_n(x)=v(nx)$: then $\|v_n\|_2 = n^{-1/2}\|v\|_2$.