Well i'd say it depends of the context but one reason that come to my mind is that the borel $\sigma$-algebra is simpler (and smaller) than the Lebesgue $\sigma$-algebra $\mathscr{M}(\mathbf{R})$. For a lot of things the seting of borel functions or borel sigma alegbra is enough for what you want to do, using the Lebesgue sigma algebra would only make the proofs harder or even invalidate the results you want to prove.
An example about the "harder proofs" parts : the $\sigma$-algebra $\mathscr{B}(\mathbf{R})$ is generated by the open sets of $\mathbf R$, and a lot of proofs use this fact. Unfortunatly the situation is more complex with $\mathscr{M}(\mathbf R)$.
An example about the "invalidating results" part : It's easy to show that if $f$ and $g$ are Borel then $f\circ g $ is also Borel. However, if you define a measurable function to be a function $f$ such that for every open set $U\subset \mathbf R$ you have $f^{-1}(U)\in \mathscr M (\mathbf R)$ then the composition of two measurable functions is not measurable in general.
Side note : the fact that the composition of a two measurable functions is not measurable is closely related to the fact that some functions are Borel but not Lebesgue (where $f$ is Lebesgue mean $f^{-1}(U)\in \mathscr M (\mathbf R)$ for every $U\in \mathscr M (\mathbf R))$. There is a exercice in Folland's Real analysis about that if i remember it right. But $\mathscr M (\mathbf R)$ is absolutely crucial in integration theory, indeed there are functions that are Riemann integrable but not Borel (think of the characteristic functions of some subset of the triadic cantor set).
To finish, yes $\mathscr M (\mathbf R)\backslash\mathscr B (\mathbf R)$ is nonempty. But you have the following result :
if $A\in \mathscr M (\mathbf R)\backslash\mathscr B (\mathbf R)$ then there exists two borel sets $M$ and $N$ such that $M\subset A$, $A\subset M \cup N$ and $\lambda(N)=0$ (so $A$ is a borel set up to some non Borel negligible set). Moreover one have $\lambda(A)=\lambda(M)$.
As a prelude, let me first say that $A\times B$ is clearly in the product $\sigma$ algebra, since the product sigma algebra $\mathcal{F}\otimes \mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e:
$\mathcal{F}\otimes \mathcal{G}= \sigma\Big( \{ E\times F: E\in \mathcal{F}, F\in \mathcal{G} \} \Big)$
Going by this definition, your set $A\times B$ is Lebesgue measurable in $\mathbb{R}^{n+m}$. However going by your criterion, recall first that for $A\in \mathcal{L}^n$ and $B\in \mathcal{L}^m$, we have:
$\lambda_n \otimes \lambda_m (A\times B)= \lambda_n(A)\times \lambda_m(B)$
Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that:
$U_1\supseteq A \supseteq V_1$ , $U_2\supseteq B \supseteq V_2$ while $\lambda_n(U_1\setminus V_1)\leq \frac{1}{2} \tilde{\epsilon}$ and $\lambda_m(U_2\setminus V_2)\leq \frac{1}{2} \tilde{\epsilon}$
Notice that $U_1\times U_2\supseteq A\times B \supseteq V_1\times V_2$ while $U_1\times U_2$ is open and $V_1 \times V_2$ is closed. Since you can write:
$U_1\times U_2= \Big( U_1\times V_2 \Big) \sqcup \Big( U_1 \times (U_2 \setminus V_2) \Big)= \Big( V_1\times V_2 \Big) \sqcup \Big( (U_1\setminus V_1)\times V_2 \Big) \sqcup \Big( U_1 \times (U_2\setminus V_2) \Big) $
Then:
$\lambda_{n+m}\Big( (U_1\times U_2) \setminus (V_1 \times V_2) \Big)= \lambda_{n+m}\Big( (U_1\setminus V_1)\times V_2
\Big) + \lambda_{n+m}\Big( U_1 \times (U_2\setminus V_2)
\Big)= $
$= \lambda_n(U_1 \setminus V_1)\cdot \lambda_m(V_2)+ \lambda_n(U)\cdot \lambda_m(U_2 \setminus V_2)$
If you assume $\lambda_n(U_1), \lambda_m(U_2)\leq M$, then for $\tilde{\epsilon}= \dfrac{\epsilon}{2M}$, you've shown that the difference is of measure less than $\epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $\sigma$-finite case.
Best Answer
Suppose first that $E$ is of finite measure. Since $E$ is measurable, for each $\varepsilon > 0$ there exists by definition $E \subseteq G$ open with $|G \setminus E| < \varepsilon$. Similarly we have $F \subseteq E$ open with $|E \setminus F| < \varepsilon$ (just observe that $E^\complement$ is measurable and use the former). Thus, for each $n \geq 1$ we have $F_n \subseteq E \subseteq G_n$ with $F_n$ closed, $G_n$ open and $|E\setminus F_n|,|G_n \setminus E| < \frac{1}{n}$. Now set $F = \bigcup_n F_n$ and $G = \bigcap_n G_n$ which are $F_\sigma$ and $G_\delta$ respectively (in particular, Borel sets), and note that
$$ |E \setminus F| \leq |E \setminus F_n| < \frac{1}{n} \to 0 $$ and $$ |G \setminus E| \leq |G_n \setminus E| < \frac{1}{n} \to 0. $$
So we have $|G \setminus F| = |G| - |F| = 0$, because $|G| = |E| = |F|$.
Finally, suppose that $E$ has infinite measure an let $E_n = E \cap \overline{B}(0,n)$. By the previous result, we have Borel sets $F_n \subseteq E_n \subseteq G_n$ with $|F_n| = |E| = |G_n|$ and so if $F = \bigcup_nF_n$, $G = \bigcup_n G_n$, we get $F \subseteq E \subseteq G$. To conclude, recall that since $F$ and $G$ are increasing union of measurable sets,
$$ |F| = \lim_n |F_n| = \lim_n|E_n| = |E| = +\infty $$
and similarly for $G$.