Why don't we define a measurable function to be inverse image of measurable set is measurable In royden's book, he defines a measurable function satisfying that inverse image of $[a, + \infty )$ is measurable. It seems to be quite restrictive, and I understand that my definition in the topic is more general. What are the drawbacks of my definition? 
 A: In this case, the author is considering measurable functions from a measure space $(X,\Sigma)$ to the real line with the Borel $\sigma$-algebra. In addition to being generated by open sets in $\mathbb{R}$, the Borel $\sigma$-algebra can be generated by intervals of the form $[a,\infty)$. 
Indeed, let $-\infty < a < b < \infty$ and define $$a_n = a + \frac{1}{n} \qquad \text{ and } \qquad b_n = b -\frac{1}{n}$$
Then 
$$ (a,b) = \bigcup_{n=1}^\infty [a_n,b_n)$$
where  $[a_n,b_n)=  [a_n,\infty ) \cap [b_n,\infty)^c$. 
A: Let me rephrase the question a little bit:
Why a function $f: R\to R$ is defined measurable only in the sense of $f: (R, L)\to (R, B),$ where $L$ is Lebesgue measure, and $B$ is Borel measure. In other words, what is the point of non-symmetry here? Why not define a measurable function in the sense of $f: (R, L)\to(R, L)$.
The answer is that to introduce integral, measurable functions in the former sense is sufficient. While measurable in the latter sense is actually more restrictive. And there would be  no extra bonus for that restriction.
The last point is only my own opinion. it is nearly impossible that one encounters a function which is measurable in the former sense but not measurable in the latter sense, unless of course someone construct  a counter example deliberately. That function must be very strange, it is not likely you need to deal with one in real life.
A: You're absolutely right. Here is how it goes..

Definition: If $(X,{\cal A})$ and $(Y,{\cal B})$ are measurable spaces, one says that a function $f\colon X\to Y$ is measurable if $f^{-1}(B)\in{\cal A}$ for every $B\in {\cal B}$.

It's well known that, if ${\cal B}$ is the $\sigma$-algebra generated by a set ${\cal S}\subseteq 2^Y$, then $f$ is measurable if and only if $f^{-1}(S)\in{\cal A}$ for every $S\in {\cal S}$.
Therefore, if $Y=\mathbb R$ and ${\cal B}$ is the Borel $\sigma$-algebra (which is generated by the set of intervals of the form $[a,\infty[$), then it is sufficient check the condition in the definition only for those intervals.
