I am following the solution for question 3 found here (the solution is on page 3):
The question reads:
"
Let $R$ be the unit square $[0, 1] \times [0, 1]$ in the plane, and let \mu be the usual Lebesgue measure on the real Cartesian plane. Let $N$ be the function that assigns to each real number $x$ in the unit interval the positive integer that indicates the first place in the decimal expansion of $x$ after the decimal point where the first $0$ occurs. If there are two expansions, use the expansion that ends in a string of zeroes. If $0$ does not occur, let $N(x) = \infty$. For example, $N(0.0) = 1, N(0.5) = 2, N(1/9) = \infty$, and $N(0.4763014 \dots) = 5$. Evaluate
$$ \int\int_Ry^{-N(x)}dA. $$"
My solution follows the given solution for the most part, but I find that, for a fixed $y$:
$$ \int_0^1 y^{-N(x)}dx=\sum_{k=1}^\infty9^{k-1}10^{-k}y^{-k}=\frac{1}{10y}\frac{1}{1-\frac{9}{10y}}=\frac{1}{10y}\frac{10y}{10y-9}=\frac{1}{10y-9}. $$
This ultimately leads to a wrong answer too, something like $\log(10y-9)|_{y=0}^1,$ which doesn't make sense.
Am I just making a dumb math error? Or am I missing something deeper? Or, is it possible the given solution is incorrect?
The given solution is incorrect.
Notice that $N(x) \geq 1$ by definition and $$ \int_0^1 \; y^{-N(x)} \, \mathrm{d}y $$ diverges if $N(x) = 1$. The (literal) first line of the solution shows $\mu(N^{-1}(1)) = 1/10$, so $$ \int \int_{[0,1/10) \times [0,1]} \; y^{-N(x)} \, \mathrm{d}y = \infty \text{.} $$ Adding more positive contributions from other values of $N$ does not reduce the result.
The error seems to start around \begin{align*} \int_0^1 \; y^{-N(x)} \,\mathrm{d}x &= \sum_{k=1}^\infty 9^k 10^{-k} y^{-k} \\ &= \frac{y}{10} \, \frac{1}{1- \frac{9}{10y}} \text{.} \end{align*}
As you observe, the power of $9$ in the sum should be $k-1$. Regardless, neither the sum shown nor the corrected version gives the final RHS. \begin{align*} \sum_{k=1}^\infty 9^k 10^{-k} y^{-k} &= \frac{1}{1-\frac{9}{10 y}} & &\left(-1 < \frac{9}{10y} < 1 \right) & \text{and} \\ \sum_{k=1}^\infty 9^{k-1} 10^{-k} y^{-k} &= \frac{1}{9} \frac{1}{1-\frac{9}{10 y}} & &\left(-1 < \frac{9}{10y} < 1 \right) \end{align*}
(There is an additional error in the presentation of the solution and in your thinking that is highlighted in the above two lines.)