I am trying to solve the following problem:
Let $h : \mathbb{R} \rightarrow \mathbb{R}$ be the periodic extension of the function \begin{equation} h(x) := \begin{cases} a & 0 \leq x < \lambda \\ b & \lambda \leq x < 1 \end{cases} \end{equation} and let $w_k : [0,1] \rightarrow \mathbb{R}$ for $k \in \mathbb{N}$ be defined by \begin{equation} w_k(x) = h(kx) \end{equation} Show that \begin{equation} w_k \rightharpoonup^* \int^{1}_{0}{h(x) dx} = \lambda a + (1-\lambda) b \mbox{ in } L^{\infty}(0,1). \end{equation} So i choose $g \in L^1(0,1)$ arbitrarly and since $ \lambda \in (0,1)$ we get: \begin{align} &\left| \int^{1}_{0}{g(x) w_k(x) dx} - \int^{1}_{0}{g(x)(\lambda a + (1-\lambda) b) dx} \right| = \left| \int^{1}_{0}{g(x) (w_k(x) - \lambda a - (1-\lambda) b) dx} \right| \\ &= \left| \int^{1}_{0}{g(x) (h(k x) - \lambda a - (1-\lambda) b) dx} \right| \\ &= \frac{1}{k} \left| \int^{k}_{0}{g(y/k) (h(y) - \lambda a - (1-\lambda) b) dy} \right| \\ &= \frac{1}{k} \left| \sum^{k-1}_{n=0}\left( \int^{1+n}_{\lambda + n}{g(y/k)(b - \lambda a - (1-\lambda)b) dy} + \int^{\lambda+n}_{n}{g(y/k)(a - \lambda a - (1-\lambda) b) dy} \right)\right| \\ &= \frac{1}{k} \left| \sum^{k-1}_{n=0}\left( \int^{1+n}_{\lambda + n}{g(y/k)( -\lambda a + \lambda b) dy} + \int^{\lambda+n}_{n}{g(y/k)( (1 - \lambda) a - (1- \lambda) b) dy} \right)\right| \\ &= \frac{1}{k} \left| \sum^{k-1}_{n=0}\left( -\lambda(a-b)\int^{1+n}_{\lambda + n}{g(y/k) dy} + ( 1- \lambda) (a-b) \int^{\lambda+n}_{n}{g(y/k) dy} \right)\right| \\ &= \frac{1}{k} \left| -\lambda (a-b)\int^{k}_{0}{g(y/k) dy} + (a-b) \sum^{k-1}_{n=0} \int^{\lambda + n}_{n}{g(y/k) dy} \right| \\ &= \left| -\lambda (a-b)\int^{1}_{0}{g(x) dx} + (a-b) \sum^{k-1}_{n=0} \int^{\frac{\lambda + n}{k}}_{\frac{n}{k}}{g(x) dx} \right| \\ &= \lambda |a-b| \left| \sum^{k-1}_{n=0} \int^{\frac{\lambda + n}{k \lambda }}_{\frac{n}{k \lambda}}{g(x \lambda ) dx} -\int^{1}_{0}{g(x) dx} \right| \\ \end{align} now I would say that for $k \rightarrow \infty$ the sum on the left in the last row convergeces to the right integral, since the intervals are getting smaller and stuff the missing parts of the interval to get $(0,1)$.
Now I am not sure if this is correct. I would be thankful if anyone can tell me if I am wrong or if I am missing something.