Uniform continuity implies existence of limit of integral

Question

Let $f: (0,1) \rightarrow \mathbb{R} $ be uniformly continuous. Prove that $$ \lim_{\epsilon \to 0} \int^{1-\epsilon}_{\epsilon}\!\!f(t)dt \in \mathbb{R}$$ Any ideas?? $f$ can be extended to a continuous function $\hat{f}$ in $[0,1]$, so an integrable one. My best guess is the integral will equal $\int^{1}_{0}\hat{f}(t)\,dt$. Am I correct?? A rigorous proof will be very much appreciated. Thanks in advance.

Answer 1

Correct.

As the extension is bounded, where $|\hat{f}(x)| \leqslant M$ on $[0,1]$, we have as $\epsilon \to 0$,

$$\left|\int_0^1\hat{f} - \int_\epsilon^{1-\epsilon} f \right| = \left|\int_0^1\hat{f} - \int_\epsilon^{1-\epsilon} \hat{f} \right|\leqslant \left|\int_0^\epsilon \hat{f}\right| + \left|\int_{1-\epsilon}^1 \hat{f} \right| \leqslant 2M\epsilon \to 0$$

Answer 2

Hint: $\int_\epsilon^{1-\epsilon} f = \int_\epsilon^{1-\epsilon} \hat f.$

Answer 3

Indeed, $f$ can be extended to $\hat f$ on $[0,1]$. Let $\epsilon > 0$. Then \begin{align} \bigg| \int_{0}^1 \hat fdx - \int_{\epsilon}^{1- \epsilon} \hat f dx\bigg| &= \bigg| \int_{0}^{\epsilon} \hat f dx + \int_{1-\epsilon}^{\epsilon} \hat f dx \bigg| \\ &\leq 2\epsilon ||\hat f||_{\infty}\end{align} where $||\hat f||_{\infty}$ is the supremum of $\hat f$ over $[0,1]$ and is finite by uniform continuity on a compact set.