Assume f element of R(alpha) on [a,b] where alpha is of bounded variation on [a,b] .let V(x) denote the total variation of alpha on [a,x] if a less than x less than or equal to b and let V(a) =0 .show that | integral a to b fdalpha| less than or equal to integral a to b |f| dv less than or equal to MV (b).
Consider,
∣∫abfd(α)∣≤∫ab∣fd(α)∣≤∫ab∣f∣∣d(α)∣≤sup∫ab∣f∣∣d(α)∣=∫ab∣f∣sup∣d(α)∣=∫ab∣f∣dv;v=sup∣α∣≤sup∫ab∣f∣dv=∫absup∣f∣dv=sup∣f∣∫abdv=MV∣ab;M=sup∣f∣[a,b]=M(V(b)−V(a))=M(V(b)−0);since,v(a)=0=MV(b)Therefore,∣∫abfd(α)∣≤∫ab∣f∣dv≤MV(b)|\int_a^bfd(\alpha)|\le \int_a^b|fd(\alpha)|\\ \le\int_a^b|f||d(\alpha)|\\ \le sup\int_a^b |f||d(\alpha)|\\ =\int_a^b |f|sup |d(\alpha)|\\ =\int_a^b|f|dv ; v= sup|\alpha| \\ \le sup \int_a^b|f| dv\\ =\int_a^bsup|f|dv\\ =sup|f| \int_a^bdv \\ =MV|_a^b ; M= sup |f|[a, b]\\ =M (V(b)-V(a))\\ =M(V(b)-0);since, v(a) =0\\ =MV(b)\\ Therefore, \\ |\int_a^bfd(\alpha)| \le \int_a^b|f|dv\le MV(b)∣∫abfd(α)∣≤∫ab∣fd(α)∣≤∫ab∣f∣∣d(α)∣≤sup∫ab∣f∣∣d(α)∣=∫ab∣f∣sup∣d(α)∣=∫ab∣f∣dv;v=sup∣α∣≤sup∫ab∣f∣dv=∫absup∣f∣dv=sup∣f∣∫abdv=MV∣ab;M=sup∣f∣[a,b]=M(V(b)−V(a))=M(V(b)−0);since,v(a)=0=MV(b)Therefore,∣∫abfd(α)∣≤∫ab∣f∣dv≤MV(b)
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