Indicatorfunction

Question

Can you help me to understand why the following equation is true? $$|e^{-x}| \cdot \mathbb{1}_{(0,1)}(1-e^{-x})=e^{-x}\mathbb{1}_{(0,\infty)}(x)$$

Thanks a lot.

Answer 1

The function on the left hand side implies that

$$f(x) = \begin{cases} |e^{-x}| &\text{if $0 \lt 1 - e^{-x} \lt 1$}\\ 0 &\text{otherwise} \end{cases}$$

Solving the inequality,

$$0 \lt 1-e^{-x} \lt 1$$ $$-1 \lt e^{-x}-1 \lt 0$$ $$0 \lt e^{-x} \lt 1$$ $$ln(0) \lt -x \lt ln(1)$$ $$ -\infty \lt -x \lt 0$$ $$ 0 \lt x \lt \infty$$

Note, that for $x \in (0,1), e^{-x} \gt 0$. So we can remove the mod sign too.

Thus, We have

$$f(x) = \begin{cases} e^{-x} &\text{if $0 \lt x \lt \infty$}\\ 0 &\text{otherwise} \end{cases}$$

Which is precisely the RHS.