Why does a Borel measurable function imply its Lebesgue measure? Borel measurable defined as:
$f: D \rightarrow\mathbb R$ is Borel measurable if $D$ is a Borel set and for each $a\in\mathbb R$, the set $\{x\in D: f(x) > a\}$ is a Borel set.
Definition of Lebesgue measurable function is:
Given a function $f: D\rightarrow\mathbb R$, defined on some domain $D$, we say that $f$ is Lebesgue measurable if $D$ is measurable and for each $a\in\mathbb R$, the set $\{x\in D: f(x) > a\}$ is measurable.
Intuitively, I think Lebesgue measure function is essentially a function with both input and output(or say domain and range) being Lebesgue measurable sets. Since preimage of a Borel set is another Borel set as well under Lebesgue measurable function and Borel measurable function asks for a Borel set, a Lebesgue measurable set, as the image and with the domain is a Borel set as well, I can claim that a Borel measurable function is Lebesgue measurable. 
I'm not sure whether my idea is correct especially which one of them, the range set and domain set, is required being Lebesgue measurable first? Hope some can help me correct it or offer me with better explains or proofs. Appreciate much^_^
update:
Range being measurable should be first.

The set of Borel sets is the smallest collection of sets that contains the open sets and is closed under countable unions and intersections and complements. The set of Lebesgue measurable sets is the smallest collection of sets that contains the open sets and that is closed under countable unions and intersections and complements and which is such that for any set of measure $0$, any subset of that set is measurable. Because Lebesgue sets have this property of measure $0$ sets, we say that the $\sigma$-algebra is complete.
So, Borel sets are Lebesgue sets, but not vice versa. Borel sets don't need a measure, they just need a topology, but Lebesgue sets need a measure to complete the $\sigma$-algebra.
Using the definitions you presented, you should be able to answer your own question now.
