I want to show that the continuous mapping theorem in general is not true for convergence in mean.
So i have to show for a function $g:\mathbb{R} \longrightarrow \mathbb{R}$ continuous a.e. the following is not true $$u_n \overset{\mathcal{L}_p}{\longrightarrow} u \Longrightarrow g(u_n) \overset{\mathcal{L}_p}{\longrightarrow} g(u)$$
I chose $u_n:=\frac{x}{n} \rightarrow0$ for $n\rightarrow\infty$ , $\Omega=[0,1]$, then $$\Vert u_n-0\Vert_{p} = \int_{0}^{1} \bigg \vert \frac{x}{n} \bigg \vert^p dx = \frac{x^{p+1}}{n^p(p+1)} \Bigg \vert_{0}^{1} = \frac{1}{n^p(p+1)} \longrightarrow 0$$ as $n \rightarrow\infty$.
The idea is now to chose $g(x):=\log(x)$ $$\Vert g(u_n)-g(u) \Vert_p = \int_{0}^{1}\bigg \vert \log \left(\frac{x}{n}\right)-\lim_{y \rightarrow0}\log(y) \bigg \vert^p dx= \int_{0}^{1}\bigg \vert \log \left(\frac{x}{n}\right)-\lim_{n \rightarrow \infty}\log\left(\frac{1}{n}\right) \bigg \vert^p dx =\\=\int_{0}^{1}\bigg \vert \log \left(x\right) \bigg \vert^p dx \ge \left(\int_{0}^{1}\bigg \vert \log \left(x\right) \bigg \vert dx \right )^{\frac{1}{p}} = 1$$
can someone check if the counter example is right ? I'm not sure if i can rewrite $$\lim_{y \rightarrow0}\log(y) \, \text{to} \lim_{n \rightarrow \infty}\log\left(\frac{1}{n}\right)$$ since the n is the same as for the sequence $u_n$.