[Math] Show that the cumulative distribution function is uniformly continuous.

functional-analysisfunctionsuniform-continuity

Let $f:\mathbb R→\mathbb R$ s.t. $\int_{-\infty}^\infty |f(x)| \, dx<\infty$ and $F:\mathbb R→\mathbb R$ defined as $F(x)=\int_{-\infty}^x f(t) \, dt$. Then how to prove that $F(x)$ is uniformly continuous?

-I have got no idea how to prove it. However, I know the definition of uniform continuity and the sufficient condition of bounded derivative to prove it. As $f(x)$ may be unbounded too, which suggests that $F'(x)=f(x)$ may not bounded in some case thereby restricting me to use the sufficient condition of uniform continuity. Kindly help.

Best Answer

It suffices to show that for any $\epsilon>0$ there exists $\delta>0$ such that $$x<y, |x-y| \leq \delta \implies \int_x^y |f(t)| \, dt < \epsilon. \tag{1}$$

Hints to prove $(1)$: Fix $\epsilon>0$.

  1. We have $$\int_x^y |f(t)| \, dt = \int_{[x,y] \cap \{|f| \geq R\}} |f(t)| \, dt + \int_{[x,y] \cap \{|f|<R\}} |f(t)| \, dt =: I_1+I_2$$ for any fixed constant $R>0$ (which will be chosen in the next step).
  2. Since $f$ is integrable, we can choose $R>0$ sufficiently large such that $$I_1 \leq \int_{|f| \geq R} |f(t)| \, dt \leq \epsilon.$$ Note that $R$ does not depend on $x$ and $y$.
  3. As $$I_2 \leq R |x-y|,$$ we get $$\int_x^y |f(t)| \, dt \leq R |x-y| + \epsilon.$$
  4. Conclude.
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