# Prove that integrable implies bounded [duplicate]

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 $\int_a^x-\epsilon\leq \int_a^xf(t)-f(a)\leq \int_a^x \epsilon$

$\int_a^x|f(t)-f(a)|\leq\epsilon|x-a|$

• The definition of integrable usually requires $f$ is bounded. I guess what you're being asked is that if $f$ is not bounded, then it cannot be integrable with your definition. – Pedro Tamaroff Jun 1 '14 at 22:29
• You say "if $f$ is integrable then it is continuous on $[a,b]$". This is not true. – Gamma Function Jun 1 '14 at 22:29
• You may want to read this question. math.stackexchange.com/questions/540646/… – user60887 Jun 1 '14 at 23:32
• "on" rather than "at" is the right preposition here. A function may be bounded on a set or integrable on a set. It may be continuous at a point or differentiable at a point. – Michael Hardy Jun 2 '14 at 2:46

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.

First of all, it certainly isn't true that $f$ is Lebesgue integrable implies $f$ is bounded. Thus I will assume by integrable we mean Riemann Integrable.

With the definition of the Riemann Integral given in, for example, Rudin's Principles of mathematical Analysis and Abbott's Understanding Analysis, a proof of this fact is unnecessary as both require that $f$ is bounded for in order for it to even enter in the definition for the Riemann integral.

In these books, the integral is only defined for bounded functions so if we have a function and we know how to integrate it, then it has to be bounded.

For references, see Definition 6.2 on page 122 of Rudin or Section 7.2 starting on page 186 in Abbot's text (page numbers, etc may differ by edition).

Edit: It has been brought to my attention by @SantiagoCanez in the comments that in Real Mathematical Analysis, Pugh defines the Riemann Integral in a way that does not definitionally require boundedness. The first theorem Pugh proves once he defines the Riemann Integral is that integrability implies boundedness.

This is Theorem 15 on page 155 in my edition. This goes to show that one must first agree on definitions. I don't know how you define the Riemann Integral, but see here for Pugh's proof that integrability implies boundedness (and scroll up for his definition of the Riemann Integral.)

Considering this, and the discussion below, certain definitions require a proof that integrability implies boundedness. For one such proof, see @ReneSchipperus's answer.

• The Riemann integral can be defined without requiring boundedness beforehand, which is then derived as a consequence. The Darboux integral, on the other hand, requires boundedness from the start. – Santiago Canez Jun 2 '14 at 2:15
• @SantiagoCanez I haven't seen such a construction. Can you point to a source? – Gamma Function Jun 2 '14 at 2:15
• Pugh's "Real Mathematical Analysis" – Santiago Canez Jun 2 '14 at 2:18
• @SantiagoCanez Ah! How interesting, Pugh defines the RI in a manner very different from other texts I am more familiar with.I shall append my answer accordingly. – Gamma Function Jun 2 '14 at 2:28
• @ReneSchipperus Yes. If the interval is $[a,b]$, you can arrange that $\lim_{x\to b^-}f(x)=+\infty$ and yet the integral is finite (since it would be in essence an improper integral). – Andrés E. Caicedo Jun 2 '14 at 6:49