Making the integral arbitrarily small Can the following integral
$$\int_{-\delta}^{\delta} |f(x+u)|du$$
be made arbitrarily small by changing the value of $\delta$? Is it true for all functions $f$? Does $f$ need to be bounded? Honestly I don't think boundedness matters, because over a finite interval, a function has a unique largest value.
It's obvious that making the interval $(-\delta, \delta)$ smaller, the value of interval will get smaller. The question is whether it can be made arbitrarily small. The book I'm reading says used this assumption in one of the proofs.
 A: I quote the characterization of a Riemann integrable function (highlighting the important terms) from here:

A $\underline{bounded}$ function on a $\underline{compact}$ interval $[a, b]$ is $\underline{Riemann \,\,integrable}$ if and only if it is $\underline{continuous \,\,almost\,\, everywhere}$ (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure).

So, in the comments I gave an example with $f(x)=\frac1{|x|}$ in $[-ε, ε]$. This function (despite having only $1$ point of discontinuity) is not $\underline{bounded}$ in $[-ε,ε]$ and hence not Riemann -integrable. So, is this a counterexample for your assetion? NO. Because, the formulation $$\int_{-δ}^{δ}f(u)du$$ has only a meaning for $\underline{Riemann \,\,integrable}$ functions. So, as long as you write this integral, you assume that this function is $\underline{bounded}$ and in order to make also sense and to be able to proceed that it has a lost an at most countable set of discontinuities. So, to set things straight: Necessary condition to be Riemann integrable is to be bounded on the compact interval. Then, yes this assertion holds. If it is not bounded then it cannot be Riemann integrable and the formulation has no point (we do not even examine the assertion in this case).
A: If the function is integrable (i.e., $\int_{-\infty}^\infty f(u) \, dx$ is finite), then you are correct:
$$\lim_{\delta\to0} \int_{-\delta}^\delta f(x+u)\, dx = 0.$$
This is called absolute continuity of the Lebesgue integral, which I suspect is far deeper a theorem than you want.
On the other hand, if your function is continuous, this is still true (because continuous functions are bounded on closed intervals). I suspect you are talking only about continuous functions because of your remark that "over a finite interval, a function has a unique largest value."
If $f$ is continuous, we know there exists a constant $M$ so that $|f(x-u)| < M$ for all $u \in [-1, 1]$. So, you can now say $\int_{-\delta}^\delta |f(x-u)| \, du \leq (2\delta)M $ if $0 < \delta < 1$. This quantity goes to 0 as $\delta \to 0$.
If the function is not integrable or continuous on the whole interval, then Stef has a counter-example in the comments: $\int_{-\delta}^{\delta} \frac{1}{x} \, dx$ does not converge for any $\delta > 0$.
