Question: The Lebesgue measure $\lambda^1$ is defined on generators $(a,b]$ of the Borel-sigma algebra as $b-a$. How does this explain $$\lambda^1(\{x\})=0$$ for all real $x$? I can not write this one-element set in the form $(a,b]$.

Answer: 
Consider the intervals $I\_n:=\left(x-\frac{1}{n},x\right]$. Clearly, $\{x\}\subseteq I\_n$ for all $n\in\mathbb{N}$. Then, by monotony
$$0\le\lambda^1(\{x\})\le\lambda^1(I\_n)=x-\left(x-\frac{1}{n}\right)=\frac{1}{n}.$$
Since this holds for every $n\in\mathbb{N}$, we must have $\lambda^1(\{x\})=0$. Now, if you want to be precise, you need to make sure that $\{x\}$ even has a Lebesgue measure, but fortunately it is in the Borel $\sigma$-algebra, because $\{x\}=\bigcap\_{n=1}^{\infty}I\_n$.
As far as I am concerned, the easiest way to prove that a set $S$ has measure zero is to prove that for each $\varepsilon\in\mathbb{R}\_{++}$, we have $\lambda^1(S)\leq\varepsilon$.

In your case, it suffices to note that $\{x\}\subseteq [x-\frac{\varepsilon}{2},x+\frac{\varepsilon}{2}]$ for each $\varepsilon\in\mathbb{R}\_{++}$ and hence
$$\lambda^1(\{x\})\leq \lambda^1([x-\frac{\varepsilon}{2},x+\frac{\varepsilon}{2}])=\varepsilon \text{ for each } \varepsilon\in\mathbb{R}\_{++}.$$