This is an intermediate step to a stats problem I have:
Show that there does not exist a $g: \mathbb{R} \to \mathbb{R}$ such that $\displaystyle\int_{0}^{1/\theta}g(x) \text{ d}x = 1$ for all $\theta > 0$. (Please note that $g$ cannot depend on $\theta$.)
I thought of assuming that $g$ is continuous, so thus $g\left(\dfrac{1}{\theta}\right) = 0$ (after taking derivatives of both sides). But I'm not sure what I can do with this.
If more context is needed, please let me know.
Let's prove that no (Lebesgue) integrable function $g:\mathbb R \to \mathbb R$ can exist.
Let $\alpha=\frac{1}{\theta}>0$. We have: \begin{align} \int^\alpha_0 g(x)\,dx = 1 \end{align} for all $\alpha>0$. Now take $1\geq\alpha>0$ then: \begin{align} \int^{1}_0 g(x)\, dx = 1 , \ \ \ \ \int^{\alpha}_0 g(x)\, dx = 1 \end{align} Substracting both from each other yields: \begin{align} \int^{1}_{\alpha} g(x)\,dx = 0 \end{align} Take $f_n(x) = \mathbf{1}_{[\frac{1}{n}, 1]}g(x)$ and this function converges pointwise to $\mathbf{1}_{[0,1]}g(x)$. Clearly $$\int_0^1 f_n(x)\,dx=\int_{1/n}^1 f_n(x)\,dx=\int_{1/n}^1 g(x)\,dx=0$$ for all $n\in\mathbb{N}$. But we have: \begin{align} 1= \int_0^1 g(x)\,dx = \int_0^{1} \lim_{n\to \infty} f_n(x)\,dx = \lim_{n\to\infty}\int^1_0 f_n(x)\,dx = 0 \end{align} by The Dominated Convergence Theorem. Hence a contradiction: no Lebesgue integrable function can exist. No Riemann integrable function can exist either since Riemann integrability implies Lebesgue integrability.