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
$$\int_a^b f=\lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}).$$