The following definition is a variant of what is called the Darboux integral and

Question

The following definition is a variant of what is called the Darboux integral and could easily seen to be equivalent to the Riemann integral of elementary calculus fame: Let f: R rightarrow R be a function with compact support and define I^D^+(f):= inf {integral t: t gt f and t step function}_D^-(f):= sup{integral t: t lt f and t step function} If D_+(f) = D_(f), wc call f Darboux integrable (which is equivalent to Riemann integrable) and the common value the Darboux integral of f. Prove that if f is Darboux integrable, then f is Lebesgue integrable, and D^+(f) = D^-(f) = integral f.

Explanation / Answer

Riemann Integral states that the area under the curve using definite integral between two given limits is the sum of all the areas with maximum of all lower bounds = sum of all the areas with minimum of all upper bounds. Here in the Darboux for f: R to R , it is defined as

D+(f) = Infinimum ( t : t f and t is step function )

D-(f) = Sup ( t : t f and t is step function)

As D+(f) =D-(f), it is a Darboux integrable and D+(f) =D-(f) = f, it is Lebesque integrable.

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