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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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up vote 3 down vote accepted

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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That is what , I want to prove, If I integrate in a set of measure zero, then the integral equals zero – Susuk Oct 19 '11 at 2:32
@Susuk See my edited answer. – Ragib Zaman Oct 19 '11 at 2:49
Oh my definition of integral, consist only over intervals , how can i use this ? im a little confused, sorry for be so stupid – Susuk Oct 19 '11 at 3:19
@Susuk You are not stupid, you are just learning from a different foundation. If you tell us the definitions you are using, then we can help you. – Ragib Zaman Oct 19 '11 at 6:31
My definition covers the integral superior and the inferior, the superior integral, is the infimum of the set of all the sums , with all the partitions, taking the maximum value on any $ [ x_i , x_{i+1} ] , similar with the inferior integral, taking the sup over .... Now we define the integral in the case, that this two coincide. And all the sums run over an interval, we are using a book called of the IMPA. – Susuk Oct 19 '11 at 13:47

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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Integrable functions are bounded almost everywhere, but need not be bounded. So careful with that... – Arturo Magidin Oct 19 '11 at 5:06
Arturo: I think that's true for Lebesgue integration, but I think a Riemann-integrable function must be bounded, even at points of discontinuity; otherwise, let x be such that f(x)=oo, then we select x to be the element in a given partition with f(x)(xi+1-xi) is necessarily unboundded. But Ill double-check. – gary Oct 19 '11 at 6:28
@Gary sorry for be so slow xD, but im just learning calculus )= , I don´t understand the bound of the integral – Susuk Oct 20 '11 at 3:22
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? – gary Oct 20 '11 at 3:44

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