# Thread: lebesgue integrability and showing limit as x approaches +/-∞ of F(x) = 0.

1. ## lebesgue integrability and showing limit as x approaches +/-∞ of F(x) = 0.

a) let f be L-integrable on R. show that F(x) = integral (from 0 to x) f(t)dt is continuous.
b)show that if F is L-integrable, then lim (as x approaches +/-∞) of F(x) = 0.

i am having trouble proving these statements. i'm not sure but i think part a) involves the property of differentiating under the integral sign which is justified by the dominated convergence theorem for lebesgue integrals. but the hypothesis of the differentiating under the integral sign property requires that the derivative of f (the integrand) exists for almost all x. i don't know if it satisfies this since the only information given is that f is L integrable. as for part b) i am stuck as well and don't know how to go about it. please help.

2. Originally Posted by oblixps
a) let f be L-integrable on R. show that F(x) = integral (from 0 to x) f(t)dt is continuous.
b)show that if F is L-integrable, then lim (as x approaches +/-∞) of F(x) = 0.

i am having trouble proving these statements. i'm not sure but i think part a) involves the property of differentiating under the integral sign which is justified by the dominated convergence theorem for lebesgue integrals. but the hypothesis of the differentiating under the integral sign property requires that the derivative of f (the integrand) exists for almost all x. i don't know if it satisfies this since the only information given is that f is L integrable. as for part b) i am stuck as well and don't know how to go about it. please help.
for a) you will want to use sequential continuity.

Let $\displaystyle x_n \to x$ and consider the function

$\displaystyle \displaystyle F(x_n)=\lim_{n \to \infty}\int f \chi_{[0,x_n]}d\lambda$

Now since $\displaystyle f(x)\chi_{[0,x_n]} \le f(x)$ and $\displaystyle f(x) \in L^1$
Just use DCT and $\displaystyle F(x_n) \to F(x)$

For b) what would happen it the function didn't go to zero?

3. Originally Posted by oblixps
a) let f be L-integrable on R. show that F(x) = integral (from 0 to x) f(t)dt is continuous.
b)show that if F is L-integrable, then lim (as x approaches +/-∞) of F(x) = 0.

i am having trouble proving these statements. i'm not sure but i think part a) involves the property of differentiating under the integral sign which is justified by the dominated convergence theorem for lebesgue integrals. but the hypothesis of the differentiating under the integral sign property requires that the derivative of f (the integrand) exists for almost all x. i don't know if it satisfies this since the only information given is that f is L integrable. as for part b) i am stuck as well and don't know how to go about it. please help.
Examples of function L-integrable on R that doesn't tend to 0 if x tends to infinity are given in...

http://www.mathhelpforum.com/math-he...tml#post597107

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by chisigma
Examples of function L-integrable on R that doesn't tend to 0 if x tends to infinity are given in...

http://www.mathhelpforum.com/math-he...tml#post597107

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Exactly...the function could do anything at all on a set of measure zero and not affect the Lebesgue integral.

5. for part b) would the reason be along these lines: since F is L-integrable it can be written as the infinite sum of the integrals of L-integrable functions and in order for the infinite sum to converge the individual integrals in the sum must approach 0 as n approaches infinite. I'm a little confused though since it is x in F(x) that is approaching infinite. i have looked at the thread you have provided me and near the bottom i saw that F(x) must approach 0 since if it didn't the integral of |F| would not be finite. although i can intuitively see that, i am trying to reconcile that with the definitions provided in my book and that is what is causing me some confusion.

my book does not use measure sets to motivate the lebesgue integral but defines it as: let f_k be a sequence of R integrable functions such that the infinite sum of the integral (-infinite, infinite) |f_k|dx < infinite, then the lebesgue integral of f = infinite sum of f_k is:
integral of f(x) dx = infinite sum of integral of f_k dx.

can you help me make sense of this problem given my book's definition? thanks in advance.

6. So, apparently condition b) indeed holds, but functions that satisfy the hypothesis are quite limited:

Assume for the moment $\displaystyle f\geq 0$ then if there exists a measurable set $\displaystyle M \subset \mathbb{R}$ such that $\displaystyle \int_{M} f >0$ then $\displaystyle \int_{[0,\infty )} f >0$ and so $\displaystyle F$ can't be integrable there (take a sequence in $\displaystyle M$ tending to infinity and apply dominated convergence to obtain a contradiction, if $\displaystyle M$ is bounded it's easier still) and from this we deduce $\displaystyle F\notin L^1(\mathbb{R})$ contradicting the assumption. We conclude $\displaystyle f=0$ a.e. In the same way we deal with the case $\displaystyle f\leq 0$.

Now, for any function $\displaystyle g$, we define $\displaystyle g^+=\max \{ g,0\}$ and $\displaystyle g^-=\min \{ g,0 \}$. It's a standard result that $\displaystyle g\in L^1(\mathbb{R})$ iff $\displaystyle g^+,g^-\in L^1(\mathbb{R})$, so we get $\displaystyle F(x)= \int_0^x f^+(t)dt + \int_0^x f^-(t)dt = F^+(x) + F^-(x)$ (this is easily seen to be the case because the integral is monotone), but $\displaystyle F^+ \in L^1(\mathbb{R})$ iff $\displaystyle f= 0$ a.e., and analogous for $\displaystyle F^-$. We therefore must have $\displaystyle f= 0$ a.e. if $\displaystyle F\in L^1(\mathbb{R})$, in which case the conclusion trivially holds.

On the other hand if you ask that $\displaystyle F+c \in L^1(\mathbb{R})$ for some constant c, then the problem is more difficult (read I don't have a proof for this case), but certainly interesting. As an example take $\displaystyle f(t)=2|t|e^{-t^2}$ and $\displaystyle F(x)=1-e^{-x^2}$, then $\displaystyle F$ is not integrable but $\displaystyle F(x)-1$ is.

Edit: There is a mistake in the argument, it only works if the hypothesis are satisfied by $\displaystyle G(x)= \int_0^x |f(t)|dt$. I'm not sure if the argument can be adapted.

7. why does $\displaystyle \int_{M} f >0$ imply $\displaystyle \int_{[0,\infty )} f >0$? also i am not sure how to obtain the contradiction. so i take a sequence of functions f_k but each of them have to be bounded by an integrable function and i'm not sure how to pick that. i also have trouble seeing how because of this F(x) is not L-integrable.

8. Are you sure (b) isn't something like, if f is Lebesgue integrable, then $\displaystyle \lim_{x\to\infty}\int_x^\infty f(t)dt=0$?

9. yes i double checked. b) is:
let f, F be as in part a. Show that if F(x) = $\displaystyle \int_0^x f(t)dt$ is L-integrable, then $\displaystyle lim_{x\to\infty} F(x) = 0$.

i wasn't sure how to type it but, the limit is actually as x approaches +/- infinite.