Verification of a proof in Measure Theory 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.
 A: 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). 
A: The set $E =(a,\infty)$ will work for any $a>0.$
A: Your proof is not correct. The error is that $E_i= \phi^{-1}(\{a_i\})$ not $f^{-1}(\{a_i\})$. And you know nothing about $m ( \phi^{-1}(\{a_i\}) )$
