Proving that an integral converges

I can't quite recall the exact question at the moment, but my question is mainly to do with the method of answering the problem.

How would you prove that an integral converges without evaluating the integral? (The function is continuous over the interval). What exactly would I have to show to prove convergence?

Also in this context what does bounded and unbounded mean? Am I right in saying that bounded means the integral converges to a finite value?

However, if talking about general integrability then bounded (on a closed interval) implies convergence because the integral is like a sum. On a open interval this is not true though e.g. $f:\mathbb{R} \rightarrow [-1,1] : f(x)=sin(x)$. Is bounded but not convergent over the reals.

Not true, for example [1,infinity) is closed and f(x)=1 is bounded but certainly not integrable. It needs to be a closed bounded set.

Are you sure this is right? What about the characteristic function of a non-measurable set?

hang on I said a closed interval e.g. [a,b] which is a closed bounded set.



Sorry perhaps I was not clear I was only talking about functions of the form: $f:[a,b] \rightarrow \mathbb{R}$ where f is a bounded function

Sorry perhaps I was not clear I was only talking about functions of the form: $f:[a,b] \rightarrow \mathbb{R}$ where f is a bounded function

Sure. So let E be a non-measurable subset of [0,1], and define f to be the characteristic function of E. f is bounded function from [0,1] to R, but isn't integrable.

(Unless I am forgetting something, at any rate...)

You may be using different notation to me but I cosnider [1,infinity) to be a closed interval. It is definately closed and it is an interval.

Sure. So let E be a non-measurable subset of [0,1], and define f to be the characteristic function of E. f is bounded function from [0,1] to R, but isn't integrable.

(Unless I am forgetting something, at any rate...)

[1, infinity) is a closed set but not a closed interval they are different things.

Not by what I've been taught.

well continuous on a (closed) interval implies bounded. Also any continuous function on a (closed) interval is integrable easily proved from the definition of the Riemann Integral.


Also in this context what does bounded and unbounded mean? Am I right in saying that bounded means the integral converges to a finite value? - This depends on what you mean by integral if you are talking about the Riemann integral then this is not true there are bounded functions (on closed intervals) which are not Riemann integrable e.g. f:[0,1]->Reals: f(x)=1 if x rational or f(x)=0 is x irrational.

However, if talking about general integrability then bounded (on a closed interval) implies convergence because the integral is like a sum. On a open interval this is not true though e.g. $f:\mathbb{R} \rightarrow [-1,1] : f(x)=sin(x)$. Is bounded but not convergent over the reals.

well a closed set is a set whose complement is open. you agree?
a closed interval is just [a,b] where if x lies in the closed interval it can take every value including a and b.

So [1, infinity) has complement (-infinity,1) union (1, infinity) and there is a theorem which states the union of two open sets is open. Hence the complement of [1, infinity) is open, so [1, infinity) is a closed set.

However if x lies inside [1, infinity) it cannot take infinity so it is not a closed interval.

well a closed set is a set whose complement is open. you agree?
a closed interval is just [a,b] where if x lies in the closed interval it can take every value including a and b.

So [1, infinity) has complement (-infinity,1) union (1, infinity) and there is a theorem which states the union of two open sets is open. Hence the complement of [1, infinity) is open, so [1, infinity) is a closed set.

However if x lies inside [1, infinity) it cannot take infinity so it is not a closed interval.

WolframAlpha explictly says that (emphasis mine):

"A closed interval is an interval that includes all of its limit points. If the endpoints of the interval are finite numbers a and b, then the interval {x:a<=x<=b} is denoted [a,b]. If one of the endpoints is +/-infty, then the interval still contains all of its limit points (although not all of its endpoints), so [a,infty) and (-infty,b] are also closed intervals, as is the interval (-infty,infty)."

[As far as I can see, this agrees with james' definition that "a closed interval as an interval which is closed.", which is also how I happen to think of this].