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70 ：１３２人目の素数さん[sage]：2016/12/30(金) 10:15:39.16 ID:qnhxLvuU - Define $X^+ := \cup \{ B : Borel subset of X ; \nu (B) \geq 0 \}$.
Given a Borel subset $B$ of $X$, we define $B^+ := B \cap X^+$.
Suppose that a Borel subset $B$ of $X$ satisfies $\mu (B) = 0$, hence $\nu (B) = 0$.
Then, we see that $B = B^+$ so that $\nu (B) = \nu (B^+) = \nu^{\prime } (B^+) = \nu^{\prime } (B) =0$.
Consequently, it follows that $\nu^{\prime \prime } (B) = \nu^{\prime} (B) - \nu (B) = 0$.
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81 ：１３２人目の素数さん[sage]：2016/12/30(金) 21:07:22.00 ID:qnhxLvuU - First, suppose that $n = 31$. Then, we can pick the pair $(x, y) = (1, 0)$. This implies that
we cannot pick the integer such as $n = 41N + 31$ with $N = 0, 1, \cdots$. Secondly, we let
$n = 31 \cdot 2 = 62$. Then, we can pick $(x, y) = (2, 0)$. This implies that we cannot pick the
integer such as $n = 41N + 62 = 41(N + 1) + 21$ with $N = 0, 1, \cdots$. Repeating this until
$n = 31 \cdot 41 = 1271$, we see that we should avoid all $n \geq 1271$. In fact, these $n$
do not satisfy the given condition since for all $n \geq 1271$
there is $N = 0, 1, \cdots$ and $M = 1, \cdots, 41$ such as $n = 41N + 31M$.
Consequently, we have only to sucsessively check $n = 1270, 1269, \cdots$ to find the maximal
$n$ which does not have the pair $(x, y)$ of nonnegative integers such that $31x + 41y = n$.
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82 ：１３２人目の素数さん[sage]：2016/12/30(金) 21:48:36.99 ID:qnhxLvuU - The answer is 1199.
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