# How do you take the derivative of a Lebesgue integral with respect to a measure?

For a measure, $\nu$, with $x,\theta\in\mathbb{R}$ and $f:\mathbb{R}\rightarrow\mathbb{R}$, is this true?

$$\frac{d}{d\theta} \int f(x \theta)\ \nu(dx) = f(x \theta)$$

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In the title you talk about derivative "with respect to a measure", but in the post, $\theta\in\mathbb{R}$ (the measure in $\nu$). Is the title inaccurate? –  Arturo Magidin Nov 14 '11 at 22:07
If $f$ is Riemann integrable and $\nu$ is the Lebesgue measure, then $\int f(x\theta)\nu(dx) = \int f(x\theta)\,dx = \frac{1}{\theta}F(x\theta)+C$, where $F$ is a primitive of $f$. If you then take the derivative with respect to $\theta$, you would get $\frac{1}{\theta}F'(x\theta)\frac{d}{d\theta}(x\theta) = \frac{x}{\theta}f(x\theta)$. So I would say "probably not". Try it with something easy, like $f(u) = u^2$. –  Arturo Magidin Nov 14 '11 at 22:11
The variable $x$ seems to be used as a dummy variable on the left hand side but not on the right. Is there a typo? –  Noah Stein Nov 14 '11 at 22:12