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Let $f(x)$ be non-negative and decreasing for $ x > 0$. Suppose that $\int_0^\infty f(x)dx < \infty$. Let $g(x) = \sum_{n=1}^\infty f(2^nx)$. How do I prove that $\int_0^\infty f(x) dx = \int_0^\infty g(x) dx$?

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The conditions imposed on $f$ mean that we can swap the sum and integral signs to get

$$\int_0^{\infty} g(x)\, dx = \sum_{n=1}^{\infty} \int_0^{\infty} f(2^nx)\, dx$$

Substituting $y=2^nx$ into the integrals on the RHS gives

$$\int_0^{\infty} g(x)\, dx = \sum_{n=1}^{\infty} \int_0^{\infty} 2^{-n}f(y)\, dy$$

We can then use linearity properties of the integral and sum on the RHS to show that this is equal to $\int_0^{\infty} f(x)\, dx$, as required.

[Hint: think geometric series]

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By Fubini for non-negative functions, we have $$\int_0^{+\infty}g(x)dx=\sum_{n=1}^{+\infty}\int_0^{+\infty}f(2^nx)dx.$$ In each integral, do the substitution $t=2^nx$, $dt=2^ndx$ to get $$\int_0^{+\infty}g(x)dx=\sum_{n=1}^{+\infty}2^{-n}\int_0^{+\infty}f(x)dx,$$ and we are done.

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