I have a monotone real function $f$ 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?
How to make sense of $\int_0^a f'(x)x dx$ for monotone, not necessarily continuous, functions $f$
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real-analysis
integration
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1There'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. – 2012-05-05
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0@GilKalai: Maybe you should interpret the integral as the [Riemann-Stieltjes](http://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral) integral $\int_0^ax\mathrm{d}f(x)$? This seems quite general to me. – 2012-05-05
1 Answers
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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.
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0yes, that's the best way to go about it. Thanks Didier. (Regarding Chris's remarks i would still be interested in the simplest direct interpretation of the integral itself which applies in the full generality.) – 2012-05-05
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0Thanks, Gil. (Nice blog, by the way... :-)) – 2012-05-05