Solution of Problem 22 in section 4 in chapter 12 of Royden “real analysis third edition “

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Here is the question :

Let $(X, \mathcal{S}, \mu)$ and $(Y, \mathcal{T}, \lambda)$ be $\sigma$-finite measure spaces. Suppose that $g: X \rightarrow \mathbb{R}$ is a $\mu-$integrable function and that $h: Y \rightarrow \mathbb{R}$ is a $\lambda-$integrable function. Define $f: X \times Y \rightarrow \mathbb{R}$ by $f(x,y) = g(x)h(y).$ Prove that $f$ is $\mu \times \lambda$ integrable and that $$\int_{X\times Y} f d(\mu \times \lambda) = (\int_{X}g d\mu) (\int_{Y} hd\lambda).$$

And here is its solution:

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My questions are:

1- Why $\chi_{A}.\chi_{B} = \chi_{A \times B} $?

2-why $f = \chi_{A \times B}$ leads to that $f$ is integrable?which theorem said this?

3-Why $f^{-}$ looks like this?

Could anyone explain this for me please?
is there a proof for this? and also $f^+$
EDIT:

This question is also problem 10 on pg.423 in Royden "real analysis fourth edition" and I prefer an answer depending on this edition of Royden

Best Answer

  1. $\chi_A \cdot \chi_B = \chi_{A \times B}$:
    writing down the definitions we have $$\chi_A (x) = \begin{cases}1 & \text{for } x \in A \\ 0 & \text{for } x \in X \setminus A \end{cases}$$ and analogously for $\chi_B (y)$ for $y \in Y$. Therefore, if $\chi_{A} (x) \cdot \chi_{B} (y) = 1$, then $x \in A$ and $y \in B$. The other way round, if $x \in X$ and $y \in Y$, we have $\chi_A (x) \cdot \chi_B (y) =1$. But that means, by the definition of $\chi_{A \times B} (x,y)$, that $\chi_A (x) \cdot \chi_B(y) = \chi_{A \times B} (x,y)$ for $(x,y) \in X\times Y$.
  2. The integrability of $f = \chi_{A \times B}$ follows from the equation$$\int f ~ \mathrm{d} (\mu \times \nu) = \mu (A) \cdot \nu(B)$$which is shown in the text. The right hand side is $< \infty$ because $\chi_A$ and $\chi_B$ are integrable by assumption.
  3. For $f^-$ look at the definition of it: $$f^- = - f \cdot \chi_{\{f < 0\}} = - g \cdot h \chi_{\{g\cdot h <0\}} = -g \cdot h \big( \chi_{\{g > 0 , h < 0\}} + \chi_{\{g <0 , h >0\}} \big) = -h \chi_{\{h < 0\}} \cdot g \chi_{\{g >0\}} + -g \chi_{\{g<0\}} \cdot h \chi_{\{h >0\}} = h^- g^+ + g^- h^+.$$ Here I freely used the equations $\chi_A \cdot \chi_{B} = \chi_{A \cap B}$ and $\chi_A + \chi_B = \chi_{A \cup B}$. Think about why these hold. That is similar to 1.
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