Proof of $f\in L^1(\mathbb R)$ then $m\{x\mid |f|=\infty \}=0$. Is my proof correct. Let $f\in L^1(\mathbb R)$. Then $f$ is finite a.e. I did the following proof and my teacher gave me a mark of $0$. What I did is : Let $E=\{x\mid |f(x)|=\infty \}$. We have that$$ \int_E|f|\leq \int_{\mathbb R}|f|$$
If $m(E)>0$ then $$\int_E|f|=\infty \cdot m(E)=\infty,$$
and thus $\int_{\mathbb R}|f|=\infty $ which is a contradiction. 
He says that $a\cdot \infty =\infty $ for $a>0$ is more a convention than something formal. After he justify that if my proof is valid, then someone could make the proof $$\infty \cdot m(E)=\int_E|f|\leq \int_{\mathbb R}|f|<\infty ,$$
and thus $m(E)=0$. Which is not completely wrong, but with rigor $\infty \cdot 0$ is undeterminated. So he didn't accepted my proof. 

I'm a bit confuse because for me it's a complete valid proof. What do you think ?
 A: Your argument can be made rigorous, but when you observe conventions for multiplication with $\infty$ it should be understood in terms of limits.
Assume for simplicity that $f \ge 0$ and that $E = \{f = \infty\}$.  For all $n \ge 1$ you have $n \chi_E \le f$ so by monotonicity of the integral
$$n \cdot m(E) = \int_{\mathbf R} n \chi_E \le \int_{\mathbf R}f$$
In the limit you obtain (formally) $\infty \cdot m(E) \le \displaystyle \int_{\mathbf R} f < \infty$, but the actual contradiction follows from the observation that if $m(E) > 0$ you may select $$n > \frac 1{m(E)} \int_{\mathbf R} f.$$
A: Firstly, since $\infty$ is not a number, $\infty\cdot |E|$ does not make sense if $|E|=0$. For example,
$$ n\cdot\frac1n=1$$
which leads to 
$$ \infty\cdot0=1.$$
Secondly, you assume $|E|>0$ to get
$$ |E|\le\frac1{\infty}\int_R|f|dx=0 $$
and hence $|E|=0$ which is against your assumption $|E|>0$.
To void this, you can define
$$ E_n=\{x\in R: |f(x)|\ge n\} $$
and then
$$ \lim E_n=E. $$
Since
$$ \int_R|f(x)|dx \ge \int_{E_n}|f(x)|dx\ge n|E_n|$$
one has
$$ |E_n|\le \frac1n \int_R|f(x)|dx.$$
So
$$ \lim |E_n|=0 $$
or
$$ |E|=0.$$
