measure zero set, as the set of discontinuities of a function, integrals

Question

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.

Answer 1

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 $0.$

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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.

Answer 2

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.