Lebesgue integrable implies measurable if a function is lebesgue integrable, does it imply that it is measurable?
(without any other assumption)
The reason why I ask this is because royden, in his book, kind of imply about a measurable function when assuming the function to be lebesgue integrable
 A: By definition a function $f$ is called Lebesgue integrable if $f$ is measurable and $$\int |f(x)| \, \mu(dx) < \infty.$$
This has to be included in the definition because otherwise the term $$ \int |f(x)| \, \mu(dx)$$ is not defined.
A: Actually, this is the converse of the following theorem which you can start from its end to answer your question:
Let $f$ be a bounded measurable function on a set of finite measure $E$. Then, $f$ is integrable over $E$. 
The proof uses the simple approximation lemma* with $\epsilon=\frac{1}{n}$ to show that there are two simple functions $\phi_n$ and $\psi_n$ defined on $E$ such that $0 \leq \phi_n \leq f \leq \psi_n$ on $E$, and $0 \leq \psi_n -\phi_n \leq \frac{1}{n}$. Using the monotonicity and linearity of the integral of simple functions, we can easily show that the lower and upper Lebesgue integrals of $f$ are equal.
For the converse, use the last line of the proof: the equality of lower and upper Lebesgue integrals of $f$ to produce a sequence of simple functions $\phi_n$ such that $\phi_n \to f$ pointwise on $E$, and $|\phi_n| \leq f$ on $E$. Therefore, by the simple approximation theorem**, $f$ is measurable. $\Box$ 
In the Royden-Fitzpartick's book; Real Anaylsis, you can find the following definitions and results:
Lower integral = $\sup \{ \int_E \phi: \phi \text{  is  simple, and } \phi \leq f \text{  on } E\}$.
Upper integral= $\inf \{ \int_E \psi: \psi \text{  is  simple, and } \psi \geq f \text{  on } E\}$.


*

*The simple approximation lemma: Let $f$ be a measurable real-valued function on $E$. Assume $f$ is bounded on $E$. Then for
each $\epsilon>0$, there are simple functions $\phi_{\epsilon}$ and
$\psi_{\epsilon}$ defined on $E$ which have the following
approximation properties:$0 \leq \phi_{\epsilon} \leq f \leq
   \psi_{\epsilon}$ on $E$, and $0 \leq \psi_{\epsilon} -\phi_{\epsilon}
   \leq \epsilon.$

*The simple approximation theorem: An extended real-valued function $f$ on a measurable set $E$  is measurable if and only if
there is a sequence $\{ \phi_n \}$ of simple functions on $E$ which
converges pointwise on $E$ to $f$ and has the property that $|\phi_n|
   \leq f$ on $E$ for all $n$.
A: In order to define integrability, the measurability is a necessary condition. For an example, let $A$ be the Vitali set in $[0,1]$ and $B=[0,1]-A$. Define $f=\chi_A -\chi_B$. Notice that $|f|=\chi_{[0.1]}$ and hence integrable. Now range of $f$ is $\{-1,1,0\}$ and $A=f^{-1}\{1\}$, $B=f^{-1}\{-1\}$ and $[0,1]^c=f^{-1}\{0\}$. If we assume $f$ to be integrable with respect to the lebesgue measure $\lambda$ then we should be able to write $$\int f d\lambda=\int_{f^{-1}\{1\}} f d\lambda+\int_{f^{-1}\{-1\}} f d\lambda$$ and hence we have $$\int f d\lambda=\lambda(A) -\lambda(B)\ .$$ But the RHS is not defined since both $A$ and $B$ are nonmeasurable wrt $\lambda$. So without the measurability condition, the integration is not defined.
