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.)
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$\begingroup$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 $\int \limits_{\Bbb R} |f| \,dm = \infty$.
$\endgroup$ 9 $\begingroup$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
$$\int_0^\infty \sin(x) dx.$$
$\endgroup$ $\begingroup$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).
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