Translation function of Lebesgue integral is continuous

Question

Let $A =(a,b)$ where $-\infty < a < b < \infty.$ Let $f$ be a bounded Lebesgue measurable function. Define $$g(x) = \int_{(a+x,b+x)} f(t) dt.$$ Verify that $g$ is continuous.

Proof Let $\epsilon >0$. Let $x \in \mathbb{R}.$ Then $$|g(x) - g(y)| \leq |\int_{(a+x,b+x)}f(t)dt - \int_{(a+y,b+y)}f(t)dt| \leq \int_{(a,b)} |f(t+x) - f(t+y)| dt.$$

I am not sure how to proceed. Since proving continuity requires $x, y$ quite near enough, if $f$ is uniformly continuous on $A$, it might be able to bound the integral. I do not sure if the boundedness of $f$ can replaced the uniform continuity (to use boundedness to bound the integral). I am not quite sure why $f$ bounded is enough for proving this.

Any suggestions ?

Answer 1

If $a+x < a+y < b+x < b+y$ then $1_{(a+x,b+x)} - 1_{(a+y,b+y)} = 1_{(a+x,a+y)} -1_{(b+x,b+y)} $.

Answer 2

Write your integral from a+x to b+x as the sum of the integral from a+x to c ,and the integral form c to b+x and prove each is continuous where c is any point in (a,b) . Now say G(x) =is the integral of f from c to b+x .then G(y)-G(x) is the integral of f from a+x to a+y which is an interval of length |x-y| so if |f(t)}|is less or equal to M then |G(y)-G(x)| is less than or equal to M|x-y| proving continuity.