When proving Picard approximation theorem, we always suppose $f(x,y)$ is continuous on a rectangle area, that is, $|x-x_0|\leqq a$, $|y-y_o|\leqq b $.
If we get $x$ from the whole real number line $\Bbb R$, then how can we prove that the sequence $\phi_n(x)$ converges consistently to $\phi(x)$, because only when it is inside a rectangle area, we can prove $$ |\phi_n(x)-\phi_{n-1}(x)| \leqq \frac{ML^{n-1}(x-x_0)^{n}}{n!} \leqq \frac{ML^{n-1}h^n}{n!}, $$ then use Weierstrass to get converge consistently. Here $h$ is $\min(b/M,a)$, $M$ is the $\max(|f(x,y)|)$ on the rectangle area.
My question is when replacing $a$ to $\infty$, how can we repeat the whole process coz now we can not use $h$ as a bound of $(x-x_0)$
$\phi_n(x)$ is Picard approximation sequence.
$\phi_n(x)=y_0+\int_{x_0}^{x}f(t,\phi_{n-1}(t))dt$
$L$ is the Lipschitz coefficient.
If you directly replace $a$ with $\infty$, then you lose control over the growth of the solution. Thus the usual strategy is to assume first that you can replace $b$ with $\infty$ while $a$ remains finite and show that the existence proof can be carried out for any finite $a$, thus any solution can be extended without bounds.
One usual assumption is that $f$ is globally Lipschitz continuous in the second argument $y$, one constant $L$ without any restriction to a rectangular area.
As you observed, the critical quantity is now $M$. However, $M$ is in this proof construction only ever used in the bound for $\phi_1$ for the start of the iteration, as in $$ |\phi_1(x)-\phi_0(x)|=|\phi_1(x)-y_0|\le\int_{x_0}^x|f(s,y_0)|\,ds\le M\,|x-x_0|. $$ As $s\mapsto f(s,y_0)$ is a continuous function, it will have a maximum $M_a$ for any finite interval $[-a,a]$, so that above upper bound holds with $M=M_a$ and the convergence of the series via Weierstrass can be proven without further restriction for $h=a$.
Proving the uniqueness of the constructed solution has now to apply some Grönwall argument, giving $$ |\phi(x)-\psi(x)|\le|\phi(x_0)-\psi(x_0)|e^{L|x-x_0|}, $$ so that with identical initial points also the solutions have to be identical on the intersection of their domains.