Characteristic function in integral

Question

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

Answer 1

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:

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