Let $m$ be the Lebesgue measure on $\Bbb R$ and $f:\Bbb R\to [0,\infty)$ be a Lebesgue integrable function. Show that $\exists $ a measurable set $E\subset [0,\infty)$ such that $m(E)\neq m(f^{-1}(E)$
My try: Assume that $\forall $ measurable sets $E\subset [0,\infty)$ such that $m(E)=m(f^{-1}(E))$
$f:\Bbb R\to [0,\infty)$ be a Lebesgue integrable function hence $\int_{\Bbb R} f<\infty$
Thus we have $\int_{\Bbb R} \phi<\infty$ for every simple function $\phi \le f$
Now $\int_{\Bbb R} \phi=\sum_ia_im(E_i)$ where $E_i=\phi^{-1}(a_i)$ and $a_i\in [0,\infty)$ forms the range of $\phi$
Also $\{a_i\}\subset [0,\infty)$ and $m(\{a_i\})=0$ hence $m (f^{-1}(E_i))=0$
Thus $\int_{\Bbb R} \phi=0$ and hence $\int_{\Bbb R} f=\sup \{\int_{\Bbb R} \phi:\phi\le f\}=0$ which can't be true .
Is the proof correct ?Are all the facts used properly.Please help.
Your proof is not correct. In fact, if it were, it would disprove what you wanted to prove. I'll give you an outline of a proof, but you should really check all the details involved.
You want to show that there exists a measurable set $E$ such that $m(E)=m(f^{-1}(E))$. We should also assume $E$ is non empty, otherwise the statement is true trivially. Since a proof by contradiction or contraposition would involve assuming that for all measurable sets $m(E)\neq m(f^{-1}(E))$ and deriving a contradiction somewhere, that might be an indication that contradiction is not the way to go.
So for a constructive proof, we want to construct such a set $E$, which is allowed to depend on $f$. Notice that for each fixed $\alpha$, the set $E=\{x\in [0,\infty) :f(x)=x+\alpha\}$ would certainly have this property! (prove it!) Can you show that this set is measurable? Can you always choose $\alpha$ so that the set is nonempty?
Edit
For your corrected question, the proof is still not correct (see the answer given below).