Question #304922
Show that the function f defined on [0,1] by f(x)= (-1)^(n-1) for 1/(n+1) < x/n ≤ 1/n where (n=1,2,3...) is integrable on [0,1]
Expert's answer
1 discontinuity points are countable, therefore it has Lebesgue measure zero
2 f(x ) is limited(-1<=f<=1)
f integrated by [0,1]
