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I am currently working on continuity of functions and I just red that the composition of two not-continuous can be continous. Currently I can't imagine why this is the case and can someone give me an example please?

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Take any non-continuous function $\mathbb R\to \mathbb R$ that takes values in the set $[0,1]$. For instance: $$ f(x)=\begin{cases} 0&\text{ if $\lfloor x\rfloor$ is even}\\ 1&\text{if $\lfloor x\rfloor$ is odd} \end{cases} $$ Now take a non-continuous function $\mathbb R\to\mathbb R$ that is constant on $[0,1]$. For instance: $$ g(x)=\begin{cases} 0&\text{ if $0\le x \le 1$}\\ 1&\text{otherwise} \end{cases} $$ The composition of these two functions will be constant and therefore continuous. For example, the composition $g\circ f$ is the constant function $0$.

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Let $f(x) = \begin{cases} 5, & x < 3 \\ 7 , & x \ge 3 \end{cases}$, then $f(f(x)) = 7$

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Often discontinuities arise from division by 0. So, for example $f(x) = g(x) = \frac{1}{x}$ gives $f \circ g = x$ which is continuous.

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    No, this would still have a discontinuity at $x=0$. – 2017-01-23
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    $f(g(x)) = \frac{1}{1/x} = x$... has a discontinuity at $x = 0$? – 2017-01-23
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    Yes. See http://math.stackexchange.com/q/2085521/276406 and related questions. The domain of a function does not change simply by algebraic manipulation; in this case it changes from being *implicit* to requiring an *explicit* statement that $x$ may not equal $0$. – 2017-01-24
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    Oh. Yeah you're right. Instead define it piecewise, $f(x) = g(x) = 0$ if $x = 0$ and $\frac{1}{x}$ otherwise--then it satisfies OP's conditions. – 2017-01-24