1
$\begingroup$

So far, I have encountered two main definitions for the Lebesgue integral of a non-negative measurable function $f$.

1) $\int_A f d \mu = \sup _{h \leq f, \space h \space simple} \{ \int_A h d \mu \}$

2) $\int_A f d \mu = \sup \{ \inf _{x_i \in E_i} \sum_{i=1}^N f(x_i) \mu(E_i) \} = \sup \{ \sum_{i=1}^N (\inf _{x_i \in E_i} f(x_i)) \mu(E_i) \}$

where in 2), the $E_i$'s form a partition of A and the sup is taken over all such partitions.

I want to prove that they are equivalent. Here is what I did.

1) $\leq$ 2) : take $h$ a simple function such that $h \leq f$. Write $h$ in its canonical form : $h = \sum_i c_i \mathcal{1}_{A_i}$. Since $h \leq f$, we have $c_i \leq f(x_i)$ for all $x_i \in A_i$. Using the definition of the Lebesgue integral for simple functions, we then have $\int_A h d \mu = \sum_i c_i \mu (A_i) \leq \sum_{i=1}^N (\inf _{x_i \in A_i} f(x_i)) \mu(A_i) \leq \sup \{ \sum_{i=1}^N (\inf _{x_i \in E_i} f(x_i)) \mu(E_i) \}$, from which we have $\sup _{h \leq f, \space h \space simple} \{ \int_A h d \mu \} \leq \sup \{ \sum_{i=1}^N (\inf _{x_i \in E_i} f(x_i)) \mu(E_i) \}$.

2) $\leq$ 1) : let $E_i$'s be a partition of A, and $\alpha = \sum_i \inf_{x_i \in E_i} \{f(x_i) \} \mu (E_i)$. Fix $\epsilon > 0$. We can choose ($x_i^*)_i$ such that $\inf f(x_i) \leq f(x_i^*) \leq \inf f(x_i) + \epsilon$. We have $\alpha \leq \sum _i f(x_i^*) \mu (E_i)$. We set $h^* = \sum _i (f(x_i^*) - \epsilon) \mathcal{1}_{E_i}$. Then $h^* = \sum _i (f(x_i^*) - \epsilon) \mathcal{1}_{E_i} \leq \sum _i (\inf_{x_i \in E_i} f(x_i)) \mathcal{1}_{E_i} \leq f$. We also have $\int_A h^* d \mu = \sum_i (f(x_i^*) - \epsilon) \mu (E_i) = \sum_i (f(x_i^*) \mu (E_i) - \epsilon \mu (A) $. So $\alpha \leq \int_A h^* d \mu + \epsilon \mu(A) \leq \sup _{h \leq f, \space h \space simple} \{ \int_A h d \mu \} + \epsilon \mu (A)$. Take $\epsilon \to 0$ so that $\alpha \leq \sup _{h \leq f, \space h \space simple} \{ \int_A h d \mu \}$ and finally $\sup \{ \sum_{i=1}^N (\inf _{x_i \in E_i} f(x_i)) \mu(E_i) \} \leq \sup _{h \leq f, h simple} \{ \int_A h d \mu \}$.

Is it correct ? Can ce generalize the second part to the case $\mu(A) = \infty$ ?

If this is correct, in which case the definition 2) is preferably used ? Thanks.

$\endgroup$
2
$\begingroup$

The two definitions are equivalent.

For the $ \Rightarrow)$ your proof is correct and easy. For the second one $ (\Leftarrow) $ you can avoid the $ \epsilon$ argument really easy:

First analize the set in which the suprememum of the expression $ \sup \{ \sum_{i=1}^{N} (\inf_{x_i \in E_i} f(x_i) \mu (E_i) ) \} $ is taken. This supremum must be taken among the set of all possible finite partitions of $A$, $i.e.$ taken over all the posible $ E_1, E_2, \cdots , E_N $ such that $A = \bigcup_{i=1}^{N} E_i $ So for any finite partition $ E_1, E_2, \cdots, E_N $ you can define a simple function, say $h$ by the formula $ h = \sum_{i=1}^N (\inf_{x_i \in E_i} f(x_i)) \cdot 1_{E_i} $. Note that $ h \leq f $.

Finally because $h$ is simple, we have $ \sum_{i=1}^N (\inf_{x_i \in E_i} f(x_i)) \mu(E_i) \leq \int f d\mu $ and if you take the supremum in the so said set of partitions you get the result.

I think the second definition could be useful to prove properties of functions $f$ which can be rewritten as $ \sup\{g(x)\} $ or $ \inf\{g(x)\} $. Recall that in general you can not change $ \inf \sup $ for $\sup \inf $ but under certain assumptions you can do it. In those scenarios it may be helpful to use this definition when working with the integral.

$\endgroup$
  • $\begingroup$ Is it easier to use definition 1) or 2) to prove that $\int_{\mathbb{N}} f d \mu = \sum_{n=0}^\infty f(n)$ where $\mu$ is the counting measure on $\mathbb{N}$ ? $\endgroup$ – Vandrin Sep 27 '16 at 9:30
  • $\begingroup$ I haven't tried it with 2) but it is relatively easy with 1). $\endgroup$ – Joaquin San Sep 27 '16 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.