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.


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.

  • $\begingroup$ 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 $\mathbb{N}$ ? $\endgroup$ Sep 27 '16 at 9:30
  • $\begingroup$ I haven't tried it with 2) but it is relatively easy with 1). $\endgroup$ Sep 27 '16 at 10:42

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