TITLE: Prove that integrable implies bounded
QUESTION [5 upvotes]: If function $f$ is integrable at $[a,b]$ prove that $f$ is bounded at $[a,b]$.
 Could you explain without using rieaman integrable?
I have did so far : 
if $f$ is integrable then it is continuous at $[a,b].$ let $a\in I=[a,b]$ so for any $\epsilon>0$ there is $\delta$  such as
$$|x-a|\leq\delta\Rightarrow|f(x)-f(a)|\leq\epsilon$$
assume $|t-a|\leq\delta$ then $-\epsilon\leq f(t)-f(a)\leq \epsilon$
so $\int_a^x-\epsilon\leq \int_a^xf(t)-f(a)\leq \int_a^x \epsilon$
$\int_a^x|f(t)-f(a)|\leq\epsilon|x-a| $

REPLY [5 votes]: Here is a proof: The integral is calculated by partitioning the interval into smaller intervals, and then in each such small interval we chose a value of the function, multiply it by the length of the small interval and then sum all of these products. If the function is unbounded then in one of the smaller intervals it will still be unbounded so we can chose a value for the function in this interval so large that the resultant sum will also be large. In other words amongst all of the approximations to the integral we have sums that are arbitrary large, thus the function is not integrable.
Also even for the Riemann integral there are integrable functions that are not continuous, in fact integrable functions are a much larger class.