Variation of Peano Existence Theorem 
Peano Theorem: Let $f$ be continuous in $\Omega=I_a\times B_b$, where $I_a=\{t:|t-t_0|\leq a\}$ and $B_b=\{x:|x-x_0|\leq b\}$. If $|f|<M$ in $\Omega$, the 
  $$x'=f(t,x) \qquad x(t_0)=x_0$$
  has at least one solution in $I_\alpha$, where $\alpha=\min{\{a,b/M\}}$.

Changing the condition $|f|<M$ by $|f|\leq M$, is it possible to obtain the same results as the theorem above?
Edit: I thought I had managed to solve the problem in the post below. But as I commented on the post itself, I think I found an error, can someone help me?
 A: With the help of user Anonymous, it was possible to solve the exercise and, with the change of the hypothesis, arrive at the same conclusion as the stated theorem.
Assuming: $a\geq b/M$
By hypothesis, we have $|f|\leq M$, thus $|f|<2M$. In this way, the Cauchy problem
$$x'=f(t,x) \qquad x(t_0)=x_0$$
satisfies the hypotheses of Peano theorem. That is, there is a solution $x(t)$ in the interval $[t_0 -b/2M,t_0+b/2M]$.
Likewise, the following Cauchy problems have a solution. (See comment below)
$$y_+'=f(t,y_+)\qquad y_+(t_1)=x(t_1) \quad\text{where } t_1=t_0+b/2M$$
$$y_-'=f(t,y_-)\qquad y_-(t_2)=x(t_2) \quad\text{where } t_2=t_0-b/2M$$
Therefore $y_+$ is defined in $[t_0,t_0+b/M$] and $y_-$ in $[t_0 - b/M, t_0]$.
$$\varphi(t) =
\left\{
 \begin{array}{ll}
  y_-(t)  & \mbox{if } t\in [t_0 - b/M, t_0 -b/2M) \\
  x(t) & \mbox{if } t\in [t_0 -b/2M,t_0+b/2M] \\
  y_+(t)  & \mbox{if } t\in (t_0+b/2M, t_0 +b/M) \\
 \end{array}
\right.$$
Thus, $\varphi$ is a solution of $x'=f(t,x), x(t_0)=x_0$ defined in the interval $I_\alpha$ where $\alpha=\min\{a,b/M\}=b/M$
Assuming: $a> b/M$
Let $\epsilon=b/a-M>0$, thus $a=b/(M+\epsilon)$. Then $\vert f\vert<M+\epsilon$, so by the  Peano theorem, there is a solution $x(t)$ over $I_\alpha$ where $\alpha=\min\{a,b/(M+\epsilon)\}=a$. But we also know $a=\min\{a,b/M\}$ since $a<b/M$, so in fact $\alpha=\min\{a,b/M\}$.
Comment: I believe that I may have made a mistake in assuming that the solution exists. In Peano's theorem, $t_0$ and $x_0$ are required to be in the domain of $f$. I was able to guarantee that $t_1$ is in $I_a$. Because taking $a'=a-b/2M$ it is concluded that $I_{a'}\subset I_{a}$ and $\min\{a',b/2M\}=b/2M$.
But the problem is ensuring that $y_+(t_1)=x(t_1) \in B_b$. Is it possible to correct this proof to obtain the conclusion of the theorem?
