Lebesgue integral, integer part x

Question

$$ \int_{0}^{\infty} 10^{-2[x]} dx $$ How to solve it? is the Lebesgue integral. I drew a graph, it is piecewise continuous. Sum of this function will converge. But I can not understand how it all right to issue. How to solve the integral?

Answer 1

Split the integral into intervals where $\lfloor x\rfloor$ is constant:

$$\int_0^\infty 10^{-2\lfloor x\rfloor}dx = \sum_{n=0}^\infty\int_{n}^{n+1} 10^{-2\lfloor x\rfloor}dx = \sum_{n=0}^\infty\int_{n}^{n+1} 10^{-2n}dx = \sum_{n=0}^\infty 10^{-2n} = \frac{100}{99}$$