Proof of Tonelli's Theorem - Stein's.

Theorem 3.2 pge 80 (Stein Real Analysis): Suppose $f$ is a nonnegative measurable funtion on $\mathbb{R}^{d_1} \times \mathbb{R}^{d_2}$ then for a.e. $y$:

1. The slice $f^y$ is measurable on $\mathbb{R}^{d_1}$.
2. The function defined by $\int_{\mathbb{R}^{d_1}} f^y(x) \, dx$ is measurable on $\mathbb{R}^{d_2}$.
3. $$\int_{ \mathbb{R}^{d_2} } \Big( \int_{ \mathbb{R}^{d_1}} f(x,y) \, dx \Big) \, dy = \int_{ \mathbb{R}^d } f(x,y) \, dx dy .$$ in the extended sense.

Stein has proven the case (Fubini's Theorem, Theorem 3.1, page 76) when "nonnegative measurable" and "measurable" are replaced by "integrable" above.

Proof of 3.2: Consider $$f_k(x,y) := \begin{cases} f(x,y) & \text{ if |(x,y) | < k and f(x,y) <k } \\ 0 & \text{ otherwise } \end{cases}$$ Each $f_k$ is integrable. By Fubini's exists $E_k \subseteq \mathbb{R}^{d_2}$, $m(E_k) = 0$ such that $f^y_k$ is integrable for $y \notin E_k$. If $E:= \bigcup E_k$ we find that $\color{blue}{ \text{$f^y(x)$is measurable for all$y \in E^c$and all$k$.}}$ Moreover $m(E)=0$. $\color{blue}{\text{ Since$f^y_k \nearrow f^y$}}$, MCT implies for $y \notin E$, that $$\int_{ \mathbb{R}^{d_1} } f_k(x,y) \, dx \nearrow \int_{\mathbb{R}^{d_1}} f(x,y) \, dx \text{ as } k \rightarrow \infty .$$

The two highlighted part doesn't make sense to me.

I believe Stein meant "$f^y_k$ is measurable." for first part.

For second part, if $f^y(x) = \infty$ for some $x$, then we do not have $f^y_k(x) \rightarrow f^y(x)$ simply by definition. Also we are not even sure if $f^y(x) < \infty$ for a.e. $x$ - so we cannot apply MCT.

What is going on?

• I think you are right. For the second part, you can modify $f_k$'s so that $f_k^y \nearrow f^y$. – user438618 Jul 22 '17 at 9:30
• How would you modify that to ensure $f_k$ is still integrable? – CL. Jul 22 '17 at 9:31
• For instance, $f_k(x, y) = \min\{f(x, y) , k\}$ if $|(x, y)| <k$, and $f_k(x, y) = 0$ otherwise. – user438618 Jul 22 '17 at 9:56
• That sounds good! So we have (i) $\int f_k \, d\mu \le \int_{Q_k} k \, d\mu < k^{d}*k < \infty \Rightarrow f_k$ is integrable. (ii) If $f^y(x) = \infty$, then $f_k^y(x) \rightarrow \infty$ for $k$ sufficiently large. (??) – CL. Jul 22 '17 at 11:36