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up vote 5 down vote favorite

I am not sure how to do this. I can prove it if I know $f$ is bounded, but otherwise I am stuck.

$f$ is integrable, prove $F(x) = \int_{-\infty}^x f(t) dt$ is uniformly continuous.

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I don't think if i'm right, but if we take $f(x)=e^x$, will be $F(t)=e^x$? Is it uniformly continuous? – matgaio May 14 '12 at 21:41
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@matgaio: but the exponential is not integrable. – Martin Argerami May 14 '12 at 21:42
Ok, you are wright. I was thinking intuitivelly: $\int_{-\infty}^xe^tdt=e^t|_{-\infty}^t$ and "$e^{-\infty}=0$"... But the hypotesis was $f$ integrable over the whole $\mathbb{R}$ – matgaio May 14 '12 at 21:44
This function is actually absolutely continuous (far away better). – checkmath May 14 '12 at 23:11

2 Answers

up vote 7 down vote accepted

Let $B_M = \{ x | |f(x)|>M\}$. Show (using the dominated convergence theorem) that $ \lim_{M\rightarrow \infty} \int |f \; 1_{B_M}| = \lim_{M\rightarrow \infty} \int_{B_M} |f| = 0$. Choose $M$ such that $\int_{B_M} |f| < \frac{\epsilon}{2}$.

Then choose $x,y$ such that $|x-y|< \frac{\epsilon}{2M}$, WLOG take $x>y$. Then bound the difference with $$|F(x)-F(y)| \leq \int_y^x | f(t) | \; dt \\ = \int_{[y,x]} | f(t) 1_{B_M^C}(t)+f(t) 1_{B_M}(t) | \; dt \\ \leq M\frac{\epsilon}{2M}+ \frac{\epsilon}{2}= \frac{\epsilon}{2}+ \frac{\epsilon}{2}= {\epsilon}$$

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Hmm, in my first solution I stated that a integrable function was bounded on a set $A$ such that $m(\mathbb{R} \backslash A) = 0$. Is this true, or can I only find arbitrarly small measures? I am playing with how to prove your statement, but I am not sure exactly what to do for it. – user24900 May 14 '12 at 21:57
My attempt is, suppose $m(B_M) > \epsilon$ for all $M$. Then we have $\int_\mathbb{R} |f(t)| dt \ge \int_{B_M} |f(t)| dt \ge \int_{B_M} M dt = M\epsilon$. For all $M$. As $M$ can be arbitrarily large, we must have $\int_\mathbb{R} |f(t)| dt = \infty$. Thus $|f(t)|$ is not integrable, so $f$ is not integrable. Does this do the trick? – user24900 May 14 '12 at 22:03
I am stuck on the last inequality on your proof. I think you are using $\int_{[y, x] \cap B_M} |f(t)| dt < \epsilon /2$, but can't values of $|f(t)|$ be arbitrarily large, so even if the set is measure small, the integral might not be small. – user24900 May 14 '12 at 23:01
My apologies, I messed up the proof. I will fix it shortly. – copper.hat May 14 '12 at 23:08
It is fixed now. – copper.hat May 14 '12 at 23:15
up vote 5 down vote

Note that for $a < b$, $F(b) - F(a) = \int_a^b f(t)\ dt$. The claim is that for any $\epsilon>0$ there is $\delta>0$ such that $\left| \int_a^b f(t)\ dt\right| < \epsilon$ whenever $a<b<a+\delta$.

Note that $\lim_{b \to -\infty} F(b) = 0$ (e.g. by the Lebesgue Dominated Convergence theorem applied to $f(x) I_{x \le -n}(x)$, where $I_A$ is the indicator function of $A$), and similarly $\lim_{b \to \infty} F(b) = L = \int_{-\infty}^\infty f(t)\ dt$, and $F$ is continuous. Take $N$ so $F(-N) < \epsilon/3$ and $F(N) > L - \epsilon/3$. Using the fact that any continuous function on a bounded interval is uniformly continuous there, take $\delta > 0$ such that $|x - y| < \delta$ with $x,y \in [-N,N]$ implies $|F(x) - F(y)| < \epsilon/3$. Then e.g. if $x < N < y$ with $y - x < \delta$, $$|F(y) - F(x)| \le |F(y) - L| + |L - F(N)| + |F(N) - F(x)| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$$ Similarly we can take care of the other cases where one or more of $x$ and $y$ is outside the interval $[-N,N]$.

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