Given that ∑ k = 1 ∞ μ 0 ( S n k ) &lt; μ ∗ ( A ) + ϵ 2 n , how do we justify the non-strict inequality ∑ k = 1 ∞ μ 0 ( S n k ) ≤ ∑ n = 1 ∞ μ ∗ ( A n ) + ϵ? Is there anything other to this than if a &lt; b, then necessarily a ≤ b?

Question

Given that       ∑          k      =      1        ∞        μ    0    (      S    n    k    )  &amp;lt;      μ    ∗    (  A  )  +      ϵ          2      n      , how do we justify the non-strict inequality       ∑          k      =      1        ∞        μ    0    (      S    n    k    )  ≤      ∑          n      =      1        ∞        μ    ∗    (      A    n    )  +  ϵ? Is there anything other to this than if   a  &amp;lt;  b, then necessarily   a  ≤  b?

boomzwamhc · Accepted Answer

Though they assume that       μ          ∗        (      A    n    )  &amp;lt;  ∞ for each   n it is necessary that the final sums are finite. If       a    n    =  1 for all n and       b    n    =  2 for all   n then       a    n    &amp;lt;      b    n   but   ∑      a    n    =  ∑      b    n  .

Yahir Tucker · Accepted Answer

If   0  ≤      a    n    &amp;lt;      b    n  , then       ∑    n        a    n    ≤      ∑    n        b    n  .Consider       a    n    =      ∑    k        μ    0    (      S    n    k    ) and       b    n    =      μ    ∗    (      A    n    )  +      ϵ          2      n      .

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