The integral of (x times the derivative of f(x)) for a monotone f.

I have a monotone real function $f(x)$ defined on [0,1] with values in [0,1]. The function need not be continuous. I need to make sense of the expression: $$F(a)=\int_0^a f'(x)x dx,$$ in the greater possible generality and in the most elementary way. What is the best way to go about it.

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There's a pretty big divergence between your two goals. The most elementary way involves interpreting $f'$ as the actual derivative of $f$, which requires $f$ to be differentiable (and in particular continuous). But this won't give the greatest possible generality. – Chris Eagle May 5 '12 at 10:45
@GilKalai: Maybe you should interpret the integral as the Riemann-Stieltjes integral $\int_0^ax\mathrm{d}f(x)$? This seems quite general to me. – Dejan Govc May 5 '12 at 18:31
Thanks, Dejan --G. – Gil Kalai May 5 '12 at 21:21

Let us pretend for a moment that $f$ is regular, then an integration by parts yields $$F(a)=\left[xf(x)\right]_{0}^a-\int_0^af(x)\mathrm dx=af(a)-\int_0^af(x)\mathrm dx.$$ Since the RHS is meaningful for every monotonic function $f$ (and still others), one can define $F(a)$ by the expression in the RHS for every such function.