A function that is bounded and measurable but not Lebesgue integrable Could you give me concrete examples about
"a function that is bounded and measurable but not Lebesgue integrable".
Royden's textbook "Real analysis" says a bounded measurable function is said to be integrable if its lower Lebesgue integrale is equal to its upper Lebesgue integral.
(I know if the domain is of finite measure, then a bounded function is Lebesgue integrable iff it is measurable, so my desired example need to be on a domain of infinite measure.)
 A: This happens exactly when the integral of the positive part and the integral of the negative part are both infinite. One nice example is
$$\int_1^\infty \frac{\sin(x)}{x} dx$$
which exists in the improper Riemann sense and not in the Lebesgue sense. A more extreme example where this is easier to prove would be
$$\int_0^\infty \sin(x) dx.$$
A: Let $f : \Bbb R \to \Bbb R$ be defined as:
$$f(x) = \begin{cases} 1 & x \in [0, \infty) \\ 0 & \text{else} \end{cases}.$$  Clearly, $f$ is measurable since $f = \chi_{[0, \infty)}$ (and $[0, \infty)$ is a Lebesgue measurable set, so its characteristic function is measurable).
Also clearly $f$ is bounded.  But $\int \limits_{\Bbb R} |f| \,dm = \infty$.
A: Analytic - Lebesgue Integrable:

Any non-zero polynomial $p(x)=\sum_{i=0}^n r_i x^i$ where $r_i\in \mathbb{R}$, is not only measurable, but it is analytic!  However, by Jensen's inequiality and the continuity of (Lebesgue) measure we have that
$$
\int_{x \in \mathbb{R}} |p(x)| dx \geq 
\left|
\int_{x \in \mathbb{R}} p(x) dx
\right| 
\geq 
\left|
\int_{x \in [0,\infty)} p(x) dx
\right| 
\geq 
\lim_{t \uparrow \infty}
\left|
\int_{x \in [0,t]} p(x) dx
\right| 
\geq 
\lim_{x \mapsto \infty}
\left|
\sum_{i=0}^{n} \frac{r_i t^{i+1}}{i+1}
\right|
=\infty,
$$
where the right-hand equality holds since $\min_{i=0,\dots,n}t{|r_i|}>0$ by hypotesis of $p$ being non-zero.  
Analytic + Bounded - Lebesgue Integrable:

If you want bounded, then simply consider the case where $r_0\neq 0$ and $r_i=0$.  
Bonus: Class $C^k$ + Bounded - Lebesgue Integrable:

Since every analytic function is $k$-dimension continuisouly differentiable then for every $k\in \mathbb{N}$ there exists a $k$-times differentiable function which is not Lebesgue integrable.  In particular, since every differentiable function is continuous and continuous function is measurable, then we get what you asked for (+bonuses).  
Bonus II: Excessively many examples

Let $g$ be such that, $M\leq |g(x)|\geq \delta>0$.  Then 
$$
\int_{x \in \mathbb{R}} |g(x)p(x)| dx \geq \delta \int_{x \in \mathbb{R}} |p(x)| dx = \infty.
$$
Note: By arguing componentwise you can extend this to any Bochner space between separable Hilbert spaces (since there is only one up to isometric linear isomorphism).  
