Two questions regarding Lebesgue measure and integral

Question

I have these two homework problems and both seem intuitively wrong and I really cannot find anyway to prove them. 1) For $f \in L^1(m)$ show $m({x: f(x) = \infty}) = 0$, ($ f $ maps to $\overline{\mathbb{R}}$) How is this possible at all? Could you not define some measurable function $f$ that maps to infinity at all but some uncountable subset of $\mathbb{R}$

2. The second question seems even more off base. For $f \in L^1(\mathbb{R}^n, m) $, define $f^{\sigma}(v_1,..., v_r) = f(\sigma_1*v_1,..., \sigma_r*v_r)$. Show $f^{\sigma}$ is Lebesgue integrable and further that $\int f^{sigma} = \frac{1}{\sigma_1}*...*\frac{1}{\sigma_r}\int f$. Shouldn't it not be the reciprocal?

Answer 1

1. If $f$ maps into $\bar{\Bbb R}$, it could be even defined constantly $\infty$. However, being in $L^1(m) $ means that $\int\lvert f\rvert\, dm<\infty$.
   Let $A:=f^{-1}(\infty)$. Then we have $$\int_A \lvert f\rvert\, dm\le\int\lvert f\rvert\, dm$$

2. You can check it for one variable first, recall from calculus that if $F'(x) =f(x) $ then $[F(cx)]'=c\cdot F'(cx)=c\cdot f(cx) $ so that $$F(cx) =c\cdot\int f(cx) \, dx$$

Answer 2

Let $E_n=(|f|>n)$ for $n\ge1$. Then $$ \lVert f\rVert_1\ge\int_{E_n}|f|\,d\mu\geq m(E_n)n $$ whence $$ m(E_n)\leq n^{-1}\lVert f\rVert_1<\infty$$ Let $n\to \infty$. So $$ 0=\lim_{n\to\infty}m(E_n)=m(\cap_n E_n)=m(|f|=\infty) $$ by measure continuity from above.