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Why is depending on $y\in\mathbb{R}$ $$\int_0^1 \frac{1}{x}\cdot\chi_{(0,x)}(y)\operatorname{d}x = \int_y^1 \frac{1}{x}\operatorname{d}x\cdot\chi_{(0,1)}(y),$$

where $\chi_A(y) =\begin{cases} 1&, y\in A \\0&, y\not\in A\end{cases}$.

Thanks in advance

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    The simpliest would be to find these integrals explicitly as functions of $y$. Conceptually - look what the values of $x$ in the first integral such that the integrand is non-zero. – 2017-02-21
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    Yes - as a function of $y$ the integral should be non-zero on $]0,y]$ and zero on $]y,1]$, or not? But why then does remain only the second part of the integral? I must be wrong.. – 2017-02-21
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    No, it's the other way around… zero on $[0,y]$ and non-zero on $]y,1]$ – 2017-02-21

1 Answers 1

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First: Let $y \not\in [0,1]$ then $\chi_{(0,x)}(y) = 0$ for all $x\in [0,1]$ hence $$\int_0^1 \frac{1}{x}\cdot\chi_{(0,x)}(y)\operatorname{d}x = 0$$

Otherwise for $y \in [0,1]$ it holds $$\chi_{(0,x)}(y) = \begin{cases} 1, & x > y \\ 0, &\text{otherwise}\end{cases}$$

So $$\int_0^1 \frac{1}{x}\cdot\chi_{(0,x)}(y)\operatorname{d}x = \int_y^1 \frac{1}{x}\operatorname{d}x$$

Summarized:

$$\int_0^1 \frac{1}{x}\cdot\chi_{(0,x)}(y)\operatorname{d}x = \begin{cases} 0, & y \not\in[0,1] \\ \int_y^1 \frac{1}{x}\operatorname{d}x, & y \in [0,1] \end{cases} = \int_y^1 \frac{1}{x}\operatorname{d}x\cdot\chi_{(0,1)}(y)$$