Let $\lambda_2^\ast$ the Lebesgue outer measure on $\Bbb R^2$.
1. Let $n\in\Bbb Z$. Note that for any $\epsilon\gt 0$ $\{b\}\times (n,n+1]\subset (b-\epsilon,b]\times (n,n+1]$ so that $\lambda_2^\ast(\{b\}\times (n,n+1])\leq \lambda_2^\ast((b-\epsilon,b]\times (n,n+1])=\epsilon.$ Since epsilon is arbitrary, this proves $\lambda_2^\ast(\{b\}\times (n,n+1])=0$ and therefore by the definition of the Lebesgue measure $\{b\}\times (n,n+1]$ is measurable and $\lambda_2(\{b\}\times (n,n+1])=0,$ for each $n\in\Bbb Z$. Therefore the set $\{b\}\times\Bbb R=\bigcup_{n\in\Bbb Z} \{b\}\times (n,n+1]\tag{1}$ has measure $0$.
Consider $E\subseteq\Bbb R$ an arbitrary set of real numbers. Since $\{b\}\times E\subseteq \{b\}\times\Bbb R$ in view of $(1)$ we conclude that $\lambda_2(\{b\}\times E)=0.$ The other part is very similar.
2. Consider $P_n=\{x_0,\ldots,x_n\}$ the partition of $[a,b]$ given by $x_0=a,\quad x_k=\frac{k}{n}(b-a),\ \text{ for } k\in\{1,\ldots,n\}.$ For each $k\in\{1,\ldots,n\}$ define $m_k=\inf f([x_{k-1},x_k])\qquad M_k=\sup f([x_{k-1},x_k]).$ Note that $[x_{k-1},x_k]\times [0,M_k]=\{x_{k-1}\}\times]0,M_k]\cup]x_{k-1},x_k]\times\{0\}\cup ]x_{k-1},x_k]\times ]0,M_k]\tag{3}$ since this union is disjoint, by $1.$ we get $\lambda_2([x_{k-1},x_k]\times [0,M_k])=\lambda_2(]x_{k-1},x_k]\times ]0,M_k])=\frac{k}{n}(b-a)\cdot M_k\quad \forall k\in\{1,\ldots,n\}. \tag{2}$ For each $n\in\Bbb N$ define $L_n=\frac{b-a}{n}\sum_{k=1}^n m_k\qquad U_n=\frac{b-a}{n}\sum_{k=1}^n M_k.$ Provided that $f$ is Riemann integrable on $[a,b]$ we have $\lim_{n\to\infty} L_n=\int_a^b f=\lim_{n\to\infty} U_n.$ Now, notice that, for each $n\in\Bbb N$ $\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}\subseteq Z_n\cup \bigcup_{k=1}^n ]x_{k-1},x_k]\times ]0,M_k],$ where, in the light of $(3)$, $\lambda_2(Z_n)=0$. So $\lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\})\leq \lambda_2^\ast\left(Z_n\cup \bigcup_{k=1}^n ]x_{k-1},x_k]\times ]0,M_k]\right)\leq U_n,$ letting $n\to\infty$ we obtain $\lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\})\leq \int_a^b f\tag{4}$
Now, fix $n\in\Bbb N$. Note that for any covering $\{]a_k,b_k]\times]c_k,d_k]\}_{k\in\Bbb N}$ we have $\bigcup_{k=1}^n [x_{k-1},x_k]\times[0,m_k]\subseteq \{(x,y):0\leq y\leq f(x),a\leq x\leq b\}\subset \bigcup_{k\in\Bbb N} ]a_k,b_k]\times]c_k,d_k]$ then by similar arguments to the given above we get $L_n\leq \sum_{k=1}^\infty (b_k-a_k)(d_k-c_k)$ so $L_n\leq \lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}).$ Since this is true for each $n\in\Bbb N$, by letting $n\to\infty$ we get $\int_a^b f\leq \lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}).$ This joined with$(4)$ says $\int_a^b f=\lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}).$