If you are talking about Lebesgue integrability, yes. For example, $f(x)=1/\sqrt{x}$ on $(0,1)$ is integrable. It is also integrable as an improper Riemann integral. But for a function on a segment $[a,b]$ to be Riemann integrable (in the proper sense) it has to be bounded.
If you are talking about Lebesgue integrability, yes. For example, $f(x)=1/\sqrt{x}$ on $(0,1)$ is integrable. It is also integrable as an improper Riemann integral. But for a function on a segment $[a,b]$ to be Riemann integrable (in the proper sense) it has to be bounded.