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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3@matgaio: but the exponential is not integrable. – Martin Argerami May 14 '12 at 21:42
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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
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Related: http://math.stackexchange.com/questions/40384/showing-uniform-continuity – Jonas Meyer May 14 '12 at 22:17
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This function is actually absolutely continuous (far away better). – checkmath May 14 '12 at 23:11
2 Answers
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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 $$|F(x)-F(y)| \leq \int_y^x | f(t) | \; dt \\ = \int_{[y,x]} | f(t) 1_{B_M^C}(t)+f(t) 1_{B_M}(t) | \; dt \\ \leq M\frac{\epsilon}{2M}+ \frac{\epsilon}{2}= \frac{\epsilon}{2}+ \frac{\epsilon}{2}= {\epsilon}$$
Pedro
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copper.hat
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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. – May 14 '12 at 21:57
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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? – May 14 '12 at 22:03
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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. – May 14 '12 at 23:01
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Sorry, I just would like to ask one question about the dominated convergence theorem. Why do we have $\int \lim |f|\chi_{B_M}=0$? It seems that we have $ \lim \chi_{B_M}=\chi_{B_\infty}$ where $B_\infty:={x: |f|=\infty}$? If we do not assume that $f$ is bounded and just assume that $f\in L_1$, can we still say this set is null set? – H.Y Duan Mar 21 '24 at 07:09
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@H.YDuan $B_\infty$ must be a null set otherwise $f$ would not be integrable. – copper.hat Mar 21 '24 at 11:03
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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 $[-N,N]$.
Robert Israel
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