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I´m learning about integrals, and I have some questions. This problem consists in proving that two integrable functions $f,g:[a,b] \to \mathbb{R}$ are such that the set $ X = \{x:f\left( x \right) \ne g\left( x \right)\} $ has measure zero, then $ \int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {g\left( x \right)dx}\;. $ It is clearly equivalent to prove that the integral $ \int\limits_a^b {h\left( x \right)dx} =0$, where $ h(x) = 0 $ for all $x \in [a, b]$ except a set of measure zero.

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Your last observation is correct. To complete your thought, denote $ Y = [a,b] \setminus X $.

Then $ \int^b_a h(x) dx = \int_Y h(x) dx + \int_X h(x) =0.$

The first integral is $0$ because in $Y$, $h(x)=0.$ The second integral is $0$ because you are integrating over a set of measure $0.$


The integral of a non-negative simple function $ s(x) = \sum_{k=1}^n a_k 1_{A_k} $ where $\bigcup A_k = X $ is defined to be $\int_X s \ d\mu = \sum_{k=1}^n a_k \mu(A_k).$ Thus the integral of every simple function over a set of measure $0$ is $0$.

For arbitrary non-negative functions, the integral is defined to be the supremum of the integrals of the simple functions which approximate it from below, so the integral is $0$ again here.

For arbitrary real valued functions, the integral is defined in terms of non-negative functions:

$ \int_X f \ d\mu = \int_X f^+ \ d\mu - \int_X f^- \ d\mu $

where $f^+, f^- $ are the positive and negative parts of $f$. Thus, those get integral $0$ as well.

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    @Susuk that sounds a lot like the Riemann integral. I thought you were studying the Lebesgue integral since you referred to sets of "measure zero", but I seemed to have been wrong. See Gary's answer below for an approach with the Riemann integral ( I don't have an in depth knowledge of the Riemann integral.)2011-10-19
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If both functions are integrable, then they are both bounded , so that their difference $h(x)$ is also bounded by, say, M. Then the value of the integral is bounded by the product $M(m(Supp(h(x))$ , where $Supp(h(x)):=x:h(x) \neq 0$ . But, by definition, this last has measure zero, so that it can be covered by a collection intervals of measure $\frac {e}{2^n} $ with total measure $e$ , so that the value of the integral is bounded above by $Me$ Take $e$ to be $\frac {e}{M}$so that $Me$=e.

There is an obvious problematic self-referential use of e in the last line, but I think the idea is clear; let me know if it is not.

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    Susuk: the bound I'm using is this: the maximum value of the function times the length of the interval. For example, the integral of $x^2$ over [0,1] is bounded above by the maximum of $x^2$ over, say, [1,3]--which is 1--times the length of the interval, which is 2. Does that help?2011-10-20