Are integrable functions always bounded? So I was going through some questions, and these two made me curious:
1) An integrable function is always bounded
2) If a function is defined on an unbounded interval, then it cannot be integrable. 
I looked through the fundamental theorem of calculus, but I just don't seem to understand the connection with the boundaries, open or closed. I did see, however, that the theorem starts with : f is continuous on an interval a,b, which I assume is bounded. 
Can somebody please explain if integrable functions can indeed be UNBOUNDED?
 A: I suppose that you are talking about the Riemann integral here. If so, yes, the concept is defined only for bounded functions defined on intervals which are closed and bounded, that is, intervals of the type $[a,b]$, with $a<b$.
If $f$ is unbounded or if the interval is unbounded, we get the so-called improper integrals. For instance, we sey that $\frac1{x^2}$ is integrable on $[1,+\infty)$, because the limit$$\lim_{M\to\infty}\int_1^M\frac{\mathrm dx}{x^2}$$exists and we define$$\int_1^{+\infty}\frac{\mathrm dx}{x^2}=\lim_{M\to\infty}\int_1^M\frac{\mathrm dx}{x^2}=1.$$But this is an extension of the concept of Riemann integral.
A: Yes, an integrable function can be unbounded.  For example, the function $1/\sqrt{x}$ on the domain (0,1] is unbounded but the integral has a finite value.
A: 1) Even with the (improper) Riemann integral, an integrable function need not be bounded.
Counter-example:
The function $f$ defined as $\;f(x)=\begin{cases}\dfrac1{\sqrt x}&\text{if }\; x>0\\f(0)=&\text{whatever you want} \end{cases}$ is integrable on $[0,1]$ and its integral is
$$\int_0^1f(x)\,\mathrm d x=\lim_{\varepsilon\to 0} \int_{\varepsilon}^1\dfrac1{\sqrt x}\,\mathrm d x=\lim_{\varepsilon\to 0}2\sqrt x\,\biggr|_\varepsilon^1=2.$$
2) The function $\dfrac 1{1+x^2}$ is defined on $\mathbf R$, and integrable on this interval:$$\int_{-\infty}^{+\infty}\dfrac {\mathrm d x}{1+x^2}=\arctan x\,\biggr|_{-\infty}^{+\infty}=\frac\pi 2-\Bigl(-\frac\pi2\Bigr)=\pi.$$
A: It depends on the definition of the Riemann integrable, and that’s why we define the improper integral without only using the Riemann sum.
Some textbooks use bounded functions in the definition of the Riemann integrable and then the Riemann sum follows, but some textbooks just use the Riemann sum to define. Actually, they are of the equivalence, and we have that if a function is Riemann integrable (without considering the improper integral) if and only if it is bounded.
$$ f \text{ is integrable on a finite closed interval} \iff f \text{ is bounded}.$$
Indeed, in the Riemann sum the term $f(\xi_i)\Delta x_i$ in the neighborhood of the unbounded place is $\infty\Delta x_i=\infty$, could be as large as we like.
