The assertion of Fubini's theorem for any integrable function what has been made in the book An Introduction to Measure and Integration by Inder K Rana is not correct. It should be the following $:$
Theorem (Fubini) $:$ Let $(X, \mathcal A, \mu)$ and $(Y,\mathcal B, \nu)$ be two complete $\sigma$-finite measure spaces. Let $(X \times Y,\mathcal A \otimes \mathcal B,\mu \times \nu)$ be the product measure space induced by $(X,\mathcal A, \mu)$ and $(Y,\mathcal B, \nu).$ Let $f \in L_1(\mu \times \nu).$ Then there exist $g \in L_1(\mu)$ and $h \in L_1(\nu)$ such that $$\int_{X \times Y} f\ d(\mu \times \nu) = \int_X g\ d\mu = \int_Y h\ d\nu.$$
Let us begin the proof from the last equality what I obtained i.e. \begin{align*} \int_{X \times Y} f(x,y)\ d(\mu \times \nu) (x,y) & = \int_X \left ( \int_{Y} f^+(x,y)\ d{\nu(y)} \right ) d{\mu}(x) - \int_X \left ( \int_{Y} f^-(x,y)\ d{\nu(y)} \right ) d{\mu}(x)\ \ \ \ {\label \equation (1)}\end{align*}
Let \begin{align*} E : & = \left \{x \in X\ \bigg |\ \int_Y f^+(x,y)\ d\nu(y) < +\infty \right \} \\ F : & = \left \{x \in X\ \bigg |\ \int_Y f^-(x,y)\ d\nu(y) < +\infty \right \} \end{align*} Since the maps $x \longmapsto \displaystyle {\int_Y} f^+(x,y)\ d\nu(y)$ and $x \longmapsto \displaystyle {\int_Y} f^-(x,y)\ d\nu(y)$ are both $\mu$-integrable it follows that $\mu (E^c) = \mu(F^c) = 0.$ Define a function $g^+ : X \longrightarrow \Bbb R$ defined by $$g^+(x) = \left ( \displaystyle {\int_Y} f^+(x,y)\ d\nu(y) \right ) \chi_E (x),\ x \in X$$ and a function $g^- : X \longrightarrow \Bbb R$ defined by $$g^-(x) = \left ( \displaystyle {\int_Y} f^-(x,y)\ d\nu(y) \right ) \chi_F (x),\ x \in X$$ Then clearly $g^+(x),g^-(x) < +\infty,\ $ for all $x \in X.$ Moreover \begin{align*} g^+(x) & = \displaystyle {\int_Y} f^+(x,y)\ d\nu(y) ,\ \text{for a.e.}\ x(\mu) \\ g^-(x) & = \displaystyle {\int_Y} f^-(x,y)\ d\nu(y) ,\ \text{for a.e.}\ x(\mu) \end{align*} Let $g : = g^+ - g^-.$ Since the maps $x \longmapsto \displaystyle {\int_Y} f^+(x,y)\ d\nu(y)$ and $x \longmapsto \displaystyle {\int_Y} f^-(x,y)\ d\nu(y)$ are both $\mu$-integrable and $(X,\mathcal A,\mu)$ is a complete measure space it follows that $g^+,g^-,g \in L_1(\mu)$ and we have the following equality \begin{align*} \int_X g^+\ d\mu & = \int_X \left (\int_Y f^+(x,y)\ d\nu(y) \right ) d\mu(x) \\ \int_X g^-\ d\mu & = \int_X \left (\int_Y f^-(x,y)\ d\nu(y) \right ) d\mu(x) \\ \int_X g\ d\mu & = \int_X g^+\ d\mu - \int_X g^-\ d\mu \end{align*} From the above three equalities it follows that $$\int_X \left (\int_Y f^+(x,y)\ d\nu(y) \right ) d\mu(x) - \int_X \left (\int_Y f^-(x,y)\ d\nu(y) \right ) d\mu(x) = \int_X g\ d\mu.$$
Now from $(1)$ it follows that $$\int_{X \times Y} f\ d(\mu \times \nu) = \int_X g\ d\mu.$$
Similarly by observing that \begin{align*} \int_{X \times Y} f(x,y)\ d(\mu \times \nu) (x,y) & = \int_Y \left ( \int_{X} f^+(x,y)\ d{\mu(x)} \right ) d{\nu}(y) - \int_Y \left ( \int_{X} f^-(x,y)\ d{\mu(x)} \right ) d{\nu}(y) \end{align*} and by exploiting the completeness of the measure space $(Y,\mathcal B,\nu)$ we can find out $h \in L_1(\nu)$ such that $$\int_{X \times Y} f\ d(\mu \times \nu) = \int_Y h\ d\nu.$$
This completes the proof.
QED
If $p < \infty$, the suggestion for proving (b) is to replace $f$ by $|f|$ and apply (a) together with Fubini's theorem.
We aren't going to use Fubini to switch the order of integration—we want to use the part of Fubini's theorem which says that if you know $\int(\int|g(x,y)|\,d\nu)\,d\mu < \infty$, then the inner integral $\int|g(x,y)|\,d\nu<\infty$ $\mu$-a.e. By (a),
$$(\int(\int |f(x,y)|\,d\nu)^p\,d\mu)^{\frac1p}\le \int(\int|f(x,y)|^p\,d\nu)^{\frac1p}\,d\mu.$$
By assumption, this last expression is finite, so use Fubini to conclude $f(x,\cdot)\in L^1(\nu)$ for a.e. $x$.
As for when $p = \infty$, the suggestion is to use monotonicity of the integral, which says that if $0\le F\le G$, then $\int F\,d\mu\le \int G\,d\mu$. Consider the inequality $|f(x,y)|\le \|f(\cdot,y)\|_{L^\infty(d\mu)}$.
Best Answer
It is because $\gamma$ is not the product measure $\mu \times \nu$. Rather, $\gamma = \overline{\mu \times \nu}$, that is, $\gamma$ is the completion of the product measure $\mu \times \nu$. The measure space you are working with is $(X \times Y, \overline{\mathcal{A} \otimes \mathcal{B}}, \overline{\mu \times \nu})$.
It is not always the case that $E \in \overline{\mathcal{A} \otimes \mathcal{B}} \implies E_y \in \mathcal{A}$. For example, take $X = Y = [0, 1]$ with Lebesgue measure $m_1$ and Lebesgue sigma-algebra $L$. Let $S \subset [0, 1]$ be a set which is not Lebesgue-measurable. Then $E = S \times \{0\}$ is Lebesgue-measurable with $m_2(E) = 0$ since $S \times \{0\} \subset [0, 1] \times \{0\} \in L \otimes L$ and $m_2([0, 1] \times \{0\}) = 0$. However, $E_{0} = S$ is not Lebesgue-measurable.