If you removed middle fourths instead of middle thirds to form a generalized Cantor set G, then what would the Lebesgue measure $m(G)$?
The geometric series that I obtained from removing middle fourths was:
$\displaystyle\frac{1}{4}+(2\cdot\frac{1}{4}\cdot\frac{3}{8})+\big(4\cdot\frac{1}{4}\cdot\big(\frac{3}{8}\big)^2\big)+\big(8\cdot\frac{1}{4}\cdot\big(\frac{3}{8}\big)^3\big)+\dots$
$=\displaystyle\frac{1}{4}+(\frac{1}{2}).(\frac{3}{8})+\big(\frac{3}{8}\big)^2+2\cdot\big(\frac{3}{8}\big)^3+4\cdot\big(\frac{3}{8}\big)^4+\dots$
$=\displaystyle\frac{\frac{1}{4}}{1-\big(2\cdot\frac{3}{8}\big)}=\frac{\frac{1}{4}}{1-\frac{6}{8}}=\frac{\frac{1}{4}}{\frac{2}{8}}=1$
Did I arrive at the answer correctly? Is my geometric series correct? Any critiques are appreciated.
Yes, your geometric series is correct and your new Cantor dust has measure zero.
You may try different middle intervals to remove and get different Cantor dusts with the same measure.
Note that while these Cantor sets have the same measure, their fractal dimensions are quite different.