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

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.

Solution 1:

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 $$|F(x)-F(y)| \leq \int_y^x | f(t) | \; dt \\ = \int_{[y,x]} | f(t) 1_{B_M^C}(t)+f(t) 1_{B_M}(t) | \; dt \\ \leq M\frac{\epsilon}{2M}+ \frac{\epsilon}{2}= \frac{\epsilon}{2}+ \frac{\epsilon}{2}= {\epsilon}$$

Solution 2:

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 $[-N,N]$.