$\int_{-1}^{f(x+1/n)}g(t)dt$ converges uniformly

Question

Suppose that f is uniformly continuous, $|f(x)|\leq1$ and g is Riemann integrable. I need to show that $h_n(x) = \int_{-1}^{f(x+1/n)}g(t)dt$ converges uniformly on the reals. I've tried the following but I'm not sure about how to continue

$|h_n(x) - h(x)| = |\int_{f(x)}^{f(x+1/n)}g(t)dt|\leq M|f(x+1/n)-f(x)|$

Is it true? and if so how shall I proceed?

Answer 1

Assuming that $M$ is an upper bound of $|g|$, what I would do now would be to say that, given $\varepsilon>0$, there is some $\delta>0$ such that $|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<\frac\varepsilon M$. And so, if $N\in\Bbb N$ is such that $\frac1N<\delta$, then, for each $n\geqslant N$,$$M\left|f\left(x+\frac1n\right)-f(x)\right|<M\frac\varepsilon M=\varepsilon.$$

Answer 2

The problem is almost solved! You only need one more step. Let

$\epsilon>0$. Then by the uniform continuity of $f$ we have that there

is a $\delta >0$ such that if $1/n<\delta$ we get $\left|f(x+(1/n))-f(x) \right|<\epsilon\,\,$ for all $x$. That implies that

$sup{\left|f(x+(1/n))-f(x) \right|}$<$\epsilon$.

Thus

sup$\left|h_{n}(x)-h(x) \right|<\epsilon $ if $1/n<\delta(\epsilon)$.

i.e. for $n>\left [1/\delta(\epsilon) \right ]+1=n_{0}(\epsilon)$ we

get

sup$\left|h_{n}(x)-h(x) \right|<\epsilon $ which is all we need!!