Show $\lim_{|a|\to\infty}\int_a^{a+\epsilon} f(u(x)) d x = 0$ I have the following problem: let $u\in L^\infty(\mathbb R) \cap L^1(\mathbb R)$ and $f\in C^\infty(\mathbb R)$ with $f(0)=0$. Show that:
$$ \lim_{|a|\to\infty}\int_a^{a+\epsilon} f(u(x)) d x = 0 \qquad\text{for all }\epsilon>0$$
Now although $u\in L^1(\mathbb R)$ does not imply $u(x)\to 0$ as $|x|\to\infty$, it does imply that the average of $u$ goes to to zero in the sense that
$$ \lim_{|a|\to\infty}\int_a^{a+\epsilon} |u(x)|dx =0 \qquad\text{for all }\epsilon>0$$
So an estimate of the form
$$ \int_a^{a+\epsilon} f(u(x)) d x \approx f\Big(\int_a^{a+\epsilon} u(x) d x\Big) $$
is needed, which seems reasonable but I can't find a rigorous proof.
 A: I found a counter-example for the original question ($u$ only in $L^1$ but not necessarily bounded): consider $f(x)=x^2$ and $u$ such that on each strip $[n,n+1]$ it has the form
$$ u(x) = \begin{cases} n:& x\in[n,n+\frac{1}{n^3}] \\ 0:& x\in (n+\frac{1}{n^3}, n+1)\end{cases} $$
Then $u\in L^1(\mathbb R)$ since $\int |u(x)| dx = \sum_{n=1}^\infty \frac{1}{n^2} < \infty$, but $\int f(u(x)) dx = \sum_{n=1}^\infty \frac{1}{n}$ diverges. 
On the other hand, if $u\in L^1(\mathbb R)$ is bounded then also $u\in L^p(\mathbb R)$ for all $p$, hence then this counter example won't work. Therefore I edited my original question since in my application case I can assume that $u$ is bounded.
One strategy that this suggests is maybe applying Taylors theorem: $f(x) = f(0) + f'(0)x+\ldots$ since then we can write the integral as:
$$\int f(u(x)) d x = \int f'(0)u(x) + \frac{1}{2}f''(0)u(x)^2 +\ldots + \mathcal o(u(x)^n) d x$$
And for each coefficient but the remainder, the integral goes to zero.
A: I think I finally found an answer with an idea similar to this thread composition of $L^{p}$ functions that even generalizes the statement:

Claim: Let $g\in L^\infty(\mathbb R)\cap L^1(\mathbb R)$ and $f\in C^1(\mathbb R)$ with $f(0)=0$. Then $$f\circ g \in 
L^\infty(\mathbb R)\cap L^1(\mathbb R)$$

Proof: Since $g\in L^\infty(\mathbb R)$ it assumes, up to a set of measure zero, only values in a compact interval $[a,b]$. Thus, since the image of a compact set under a continuous map is compact, $f$ is also essentially bounded. I.e.
$$ \|f\circ g\|_{\infty} \le \max_{x\in [a,b]} |f(x)| $$
Moreover since $f$ is $C^1$, it is Lipschitz continuous on this interval, hence there is some $C>0$ such that  for all $x\in[a,b]$ holds $|f(x)|\le C |x|$. Therefore:
\begin{align}
  \|f\circ g\|_{1} = \int_{-\infty}^{+\infty} |f(g(x))| d x 
  \le \int_{-\infty}^{+\infty}C|g(x)| dx = C \|g\|_{1}
\end{align}
Which finishes the proof. The statement of the original question follows immmediately.
