Proof of $f\in L^1(\mathbb R)$ then $m\{x\mid |f|=\infty \}=0$. Is my proof correct.

Question

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.

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I'm a bit confuse because for me it's a complete valid proof. What do you think ?

Answer 1

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 $$ |E|=0.$$

Answer 2

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 $$n > \frac 1{m(E)} \int_{\mathbf R} f.$$