Suppose f is integrable on \mathbb{R}^d. Then \forall \epsilon > 0 there is a \delta > 0 such that: m(E) < \delta \hspace{0.25cm} \Rightarrow \int_E |f| < \epsilonNow, simply make the observation that: |F(x) - F(y)| = \Big| \int_{-\infty}^x f(t) dt - \int_{-\infty}^y f(t) dt \Big| \leq \ldots \ldots \leq \Big| \int_x^y f(t) dt\Big| \leq \int_x^y |f(t)|dt Since we already have that f is integrable, certainly, \forall \epsilon > 0 \hspace{0.25cm} \exists \delta > 0 such that: |x-y| < \delta \Rightarrow \int_x^y |f(t)|dt < \epsilon Thus, we have F(x) must be uniformly continuous.
In preparation for a qualifying exam in Real Analysis, during the summer of 2013, I plan to solve as many problems from Stein & Shakarchi's Real Analysis text as I can. Please feel free to comment or correct me as I make my way through this.
Thursday, June 13, 2013
2.8
Recall from Proposition 1.12 (ii) on P.65:
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