2
$\begingroup$

Problem. Suppose $-\infty\le a<b\le\infty$ and $F:(a,b)\rightarrow R$ is a $C^1$ function such that$$\int_a^b|F'(x)|dx<\infty.$$ Show that $\lim_{x\rightarrow a}F(x)$ and $\lim_{x\rightarrow b}F(x)$ exist and are finite.

[Hint. Use Dominated Convergence Theorem]


Attempt. For any $a',b'$ such that $a<a'<b'<b$, using the first fundamental theorem of calculus $$\int_{a'}^{b'}F'(x)dx=F(b')-F(a')$$ whenever either side is defined. I want to show that as $b'\uparrow b$, $\int_{a'}^bF'(x)dx+F(a')$ exists finitely.

How do I proceed?

$\endgroup$
5
  • $\begingroup$ Since $\mathbb{R}$ is complete, it suffices to establish the following Cauchy criterion: $$\lim_{s,t\uparrow b}|F(t)-F(s)| = 0.$$ In doing so, note that $$|F(t)-F(s)| = \left|\int_{s}^{t}F'(x)\,\mathrm{d}x\right|\leq\int_{\min\{s,t\}}^{\max\{s,t\}}|F'(x)|\,\mathrm{d}x.$$ Now can you relate this to the Cauchy criterion for the convergence of $\int_{c}^{b^-}|F'(x)|\,\mathrm{d}x$ for some (and in fact, any) $c\in(a, b)$? $\endgroup$ Apr 20 at 18:28
  • $\begingroup$ @SangchulLee Is this applicable to lebesgue integrals? Could you kindly elaborate/post an answer $\endgroup$
    – reyna
    Apr 20 at 19:14
  • 2
    $\begingroup$ Apply the DCT to $F(y) = F(a)+\int_a^b 1_{[a,y]}(x) F'(x) dx$. Similarly for the other. $\endgroup$
    – copper.hat
    Apr 20 at 19:27
  • $\begingroup$ @copper.hat Can you please post the elaboration? I'm unable to proceed $\endgroup$
    – reyna
    Apr 21 at 7:00
  • 1
    $\begingroup$ @reyna I added an answer. $\endgroup$
    – copper.hat
    Apr 21 at 17:37

1 Answer 1

1
$\begingroup$

Suppose $y \in (a,b)$ and choose $y' \in (a,y)$. Then $F(y) = F(Y') + \int_{y'}^y F'(t)dt = \int 1_{[y',y]}(t) F'(t)dt$

Let $y_n \to b$ (with $y_n \in (a,b)$, of course), then since $|1_{[y',y_n]}(t) F'(t)| \le |F'(t)|$, the latter is integrable, and $1_{[y',y_n]}(t) \to 1$ for all $t \in [y',b)$ we see that $F(y_n) \to F(y')+\int_{y'}^b F'(t)dt$. Since this is true for any suitable sequence we see that the limit exists.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .