I can show that any interval of the form $(a,\infty)$ is Lebesgue measurable from first principles. Thus $(0,\infty)$ and
$$ (-\infty, 1)=\bigcup_{k=1}^\infty\left(\infty,1-\frac{1}{k}\right]=\bigcup_{k=1}^\infty\left(1-\frac{1}{k},\infty\right)^C $$
are Lebesgue measurable so that
$$ (0,1)=(-\infty,1)\cap(0,\infty) $$
is Lebesgue measurable and I'm done.
I was wondering if there is a different/more direct approach to take? For example, could I start directly with the Caratheodory criterion and argue that for any set $A$ that $m^*(A) \ge m^*(A\cap(0,1))+m*(A\cap(0,1)^C)$ ?
I haven't had much luck trying this on my own and I haven't been able to find much looking online so I thought I would open the question to our community.
There is a direct approach that works because intervals have the nice property that intersection of intervals is an interval (or empty).
If $m^*A = \infty$ then there is nothing to do, so suppose $m^*A < \infty$, let $\epsilon>0$ and let $I_k$ be a collection of open intervals so that $A \subset \cup_k I_k$ and $m^*A > \sum_k l(I_k) - {1 \over 2}\epsilon$.
Let $H_k = I_k \cap (0,1)$, $J_k = I_k \cap (-\infty, {1 \over 2^{k+2}} \epsilon)$, and $L_k = I_k \cap (1-{1 \over 2^{k+2}} \epsilon, \infty)$. All of these are open intervals, and $H_k$ forms a cover of $(0,1) \cap A$ and the $J_k,L_k$ form an open cover of $(0,1)^c \cap A$.
We see that $l(H_k )+l( J_k ) + l( L_k) \le l(I_k) + {1 \over 2^{k+1} } \epsilon$, so \begin{eqnarray} m^*(A \cap (0,1)) + m^* ( A \cap (0,1)^c) &\le &\sum_k l(H_k) + \sum_k (l(J_k)+l(L_k)) \\ &\le & \sum_k l(I_k) + {1 \over 2} \epsilon \\ &\le& m^* A + \epsilon \end{eqnarray} Since $\epsilon>0$ was arbitrary, we have the desired result.