Answer to Question #114843 in Real Analysis for Sheela John

Question

Let f element R(alpha) on [a,b] where alpha is of bounded variation on [a,b] and let v(x) denote the total variation of f on [a,x] for each x in [a,b] and let v(a) =0 , show that | integral a to b f d alpha |less than or equal to integral a to b |f| dv less than or equal to M.v(b) where M is an upper bound for |f| on [a,b]

Answer 1

Now we have $f\in\mathcal R(V)$ on $[a,b]$, hence $|f|\in\mathcal R(V)$

Also $|\Delta\alpha_i|\leqslant \Delta V_i$ and we obtain

$\bigg| \sum\limits_i f(t_i) \Delta \alpha_i \bigg|\leqslant \sum\limits_i |f(t_i)||\Delta \alpha_i|\leqslant \sum\limits_i |f(t_i)|\Delta V_i$

and for the limits we have

$\bigg| \int\limits_a^b f \,d\alpha \bigg|\leqslant \int\limits_a^b|f|\,dV$

Next, $V$ increases, so for any upper Stieltjes sum $US(|f|,V)$ we have $US\leqslant M|V(b)-V(a)|=MV(b).$ Hence, we have the same for the infinum of sums, i.e. for $\int\limits_a^b|f|\,dV$ .

So, we showed that

$\bigg| \int\limits_a^b f \,d\alpha \bigg|\leqslant \int\limits_a^b|f|\,dV\leqslant MV(b)$

where $M=\sup\limits_{[a,b]} |f|$