Weak convergence of oscillating functions in $L^1(0,1)$ Let $I:=(0,1)$.
Let $f\in L^1_{loc}(\mathbb R)$ be a periodic function with period $1$. Assume 
$f|_I \in L^1(I)$ and $\int_I f(x) dx=0$. 
Define the oscillating sequence
$$
f_n(x):=f(nx).
$$
This sequence is bounded in $L^1(I)$ since $\|f_n\|_{L^1(I)} = \|f\|_{L^1(I)}$ due to periodicity.
Then it is well-known that $(f_n)$ converges weakly to zero in $L^p(I)$ for all $1<p<\infty$. It converges weakly-star to zero in $L^\infty(I) = L^1(I)^*$ and
in $L^1 \subset M(I) = C(\bar I)^*$.
The proofs of these statements use density of characteristic functions of intervalls (1) or of continuous functions (2) in the respective dual spaces. If $f\in L^p_{loc}(\mathbb R)$ for some $p>1$ then weak convergence in $L^1(I)$ follows from embedding.
However, these techniques do not work to show weak convergence in $L^1$, as its dual space is $L^\infty$, and these density arguments cease to work.
Another possibility is to use Dunford-Pettis theorem by showing that the sequence $(f_n)$ is uniformly integrable. But I do not see how to achieve this.
My question is: Is it possible to prove $f_n \rightharpoonup 0$ in $L^1(I)$ or is there a counterexample of a periodic function $f$, where this fails?
 A: As you noticed, it suffices to show that the sequence is uniformly integrable. There are several equivalent formulations of this. Since (as you noted) the sequence $(f_n)_{n \in \Bbb{N}}$ is bounded in $L^1$, it suffices to prove that $\sup_n \int_0^1 |f_n(x)| \cdot 1_{|f_n(x)| \geq M} \, d x \to 0$ as $M \to \infty$.
That this is indeed satisfied can be verified as follows:
\begin{align*}
& \int_0^1 |f_n (x)| \cdot 1_{|f_n(x)| \geq M} \, d x \\
& = \frac{1}{n} \int_0^1 n \cdot |f(n x)| \cdot 1_{|f(nx)| \geq M} \, d x \\
& = \frac{1}{n} \int_0^n |f(y)| \cdot 1_{|f(y)| \geq M} \, d y \\
& = \frac{1}{n} \sum_{i=0}^{n-1} \int_0^1 |f(y+i)| \cdot 1_{|f(y+i)| \geq M} \, d y \\
& \overset{(\ast)}{=} \frac{1}{n} \sum_{i=0}^{n-1} \int_0^1 |f(z)| \cdot 1_{|f(z)| \geq M} \, d z \\
& = \int_0^1 |f(z)| \cdot 1_{|f(z)| \geq M} \, d z.
\end{align*}
Here, we used the periodicity of $f$ at the step marked with $(\ast)$.
Note that the right-hand side of the above estimate is independent of $n$, and converges to zero as $M \to \infty$.
