Notation)
- $\mathbf{x} = (x_1, \cdots, x_n)$, $\mathbf{y} = (y_1, \cdots, y_m)$,
- $I_1=\{\mathbf{x}: a_i\le x_i\le b_i, ~~i=1, \cdots, n\}$
- $I_2=\{\mathbf{y}: c_j\le y_j\le d_j, ~~j=1, \cdots, m\}$,
- $\displaystyle L(E)=\left\{f:\int_E f \quad\text{is finite}\right\}$
- $L(d\mathbf{x})=L(I_1)$, $L(d\mathbf{y})=L(I_2)$
Theorem $6.1$ (Fubini's Theorem)
Let $f(\mathbf{x}, \mathbf{y})\in L(I), I=I_1\times I_2.$ Then
- For almost every $\mathbf{x}\in I_1$, $f(\mathbf{x}, \mathbf{y})$ is measurable and integrable on $I_2$ as a function of $\mathbf{y}$
- As a function of $\mathbf{x}$, $\int_{I_2} f(\mathbf{x}, \mathbf{y}) d\mathbf{y}$ is measurable and integrable on $I_1$, and $$ \iint_I f(\mathbf{x}, \mathbf{y}) d\mathbf{x}d\mathbf{y} = \int_{I_1} \left[ \int_{I_2} f(\mathbf{x}, \mathbf{y}) d\mathbf{y} \right] d\mathbf{x} $$
My textbook says: In these lemmas, we say that a function $f$ in $L(d\mathbf{x}d\mathbf{y})$ for which Fubini's theorem is true has property $\mathfrak{F}$.
Lemma $6.2$
A finite linear combination of functions with property $\mathfrak{F}$ has property $\mathfrak{F}$.
proof) This follows immediately from Theorems $4.9$ and $5.28$.
Theorem $4.9$ $$f \text{ and } g \text{ are measurable.}\Longrightarrow f+g \text{ is measurable.}$$
Theorem $5.28$ $$f,g\in L(E) \Longrightarrow \cases{f+g\in L(E) \\ \\ \displaystyle\int_E (f+g) = \int_E f + \int_E g}$$
Now, my question is
$1$. What does property $\mathfrak{F}$ exactly mean? I failed to analyze the following sentense: "we say that a function $f$ in $L(d\mathbf{x}d\mathbf{y})$ for which Fubini's theorem is true has property $\mathfrak{F}$." Is $f$ said to have property $\mathfrak{F}$ if $f\in L(d\mathbf{x}d\mathbf{y})$?
$2$. If so, in order to prove lemma $6.2$, theorem $4.9$ does not seem to be needed. Only theorem $5.28$ is enough to prove lemma $6.2$, isn't it?
p.s. I don't know which tags I should take. Heretofore, I tagged real analysis, lebesgue integral, lebesgue measure, measure theory, and so on. However, people often edited my tags. So, I just tagged Lebesgue Integral in this post.
I may be misreading what you've written, but I believe that the property is simply: If the results of Fubini's Theorem holds for f, then f is said to have that property.
I believe the author is just using it for notational convenience. Which text are you working with, that might help us confirm this.