Let $U = \{ f\in L^2(\sigma) \vert f = const \}$
and
$w= \frac{ 1 }{ \left\lvert \sigma \right\rvert }\int_{ \sigma }{ u(x) dx }$.
I need to proof, that $w$ is the best approximation in the sense that $$\inf_{v\in U} \left\lvert \left\lvert u-v \right\rvert \right\rvert_{L^2} = \left\lvert \left\lvert u-w \right\rvert \right\rvert_{L^2}$$ holds.
Thanks in advance for any help!
As it is suggested, I assume $0<\vert\sigma\vert<\infty$ and $u\in L^{2}(\sigma)$. Any function in $U$ is constant a.e. Hence, we have:
$$\inf_{v\in U}\Vert u-v\Vert_{L^{2}}=\inf_{c\in\Bbb R}\Vert u-c\Vert_{L^{2}}$$
Notice that minimizing $\Vert u-c\Vert_{L^{2}}$ or $\Vert u-c\Vert_{L^{2}}^{2}$ is equivalent.
$$\Vert u-c\Vert_{L^{2}}^{2}=\int_{\sigma}\vert u(x)-c\vert^{2}\text{d}x=\int_{\sigma}\vert u(x)\vert^{2}\text{d}x+c^{2}\vert\sigma\vert-2c\int_{\sigma}u(x)\text{d}x$$
As the function $u$ is fixed, minimizing the previous expression is equivalent to minimize, in terms of $c$, the following expression:
$$c^{2}\vert\sigma\vert-2\alpha c+\beta \tag{1}$$
where $\alpha=\int_{\sigma}u(x)\text{d}x$ and $\beta=\int_{\sigma}\vert u(x)\vert^{2}\text{d}x$ are constant. It is clear that $(1)$ is a parabola and thus its minimum is attained at (easily seen by differentiating $(1)$ w.r.t. $c$):
$$c=\frac{-(-2\alpha)}{2\vert\sigma\vert}=\frac{1}{\vert\sigma\vert}\int_{\sigma}u(x)\text{d}x$$
as it was claimed. Since $c=\frac{1}{\vert\sigma\vert}\int_{\sigma}u(x)\text{d}x$ is indeed a constant function, this is actually a minimum, i.e.:
$$\inf_{v\in U}\Vert u-v\Vert_{L^{2}}=\inf_{c\in\Bbb R}\Vert u-c\Vert_{L^{2}}=\min_{v\in U}\Vert u-v\Vert_{L^{2}}=\Vert u-w\Vert_{L^{2}}$$
with $$w=\frac{1}{\vert\sigma\vert}\int_{\sigma}u(x)\text{d}x$$