TITLE: Lebesgue integrable implies measurable
QUESTION [8 upvotes]: 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

REPLY [4 votes]: 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 $n$.