$ \text{Given: f is uniformly continuous on } \mathbb{R}, \ h(x) = \int_{x-1}^{x+1} f(t)dt $
$ \text{Show: h is uniformly continuous on } \mathbb{R} $
$ \text{Proof: Let } \epsilon>0, \ \delta \text{ be the radius of uniform continuity of f given } \epsilon. $
$ \Rightarrow h(x+\delta) = \int_{x-1}^{x+1} f(t-\delta)dt \text{ after a change of variables.} $
$ \Rightarrow |h(x+\delta)-h(x)| \leq \int_{x-1}^{x+1} |f(t)-f(t-\delta)|dt < 2\epsilon $
