3
$\begingroup$

Setup:

Let $(M, \mathscr A, \mu)$ be a measure space.

  • $u$ is a non-negative simple function if there exist non-negative real numbers $\alpha_1,\ldots,\alpha_p$ and measurable sets $A_1,\ldots,A_p$ such that $u=\sum_{i=1}^p \alpha_i 1_{A_i}$.
  • The integral of a non-negative simple function $u$ is $\int u\, d\mu = \sum_{i=1}^p \alpha_i \mu(A_i)$.

My question is about the definition of the integral for non-negative measurable functions. My professor defines it like this:

Let $f: M \rightarrow [0,\infty]$ be an $\mathscr A$ measurable function and $(u_n)$ an increasing sequence of non-negative simple functions which converges pointwise to $f$: $u_n \uparrow f$. Then the integral of $f$ is defined by $$\int f \, d\mu = \lim_{n\rightarrow \infty} \int u_n \, d\mu$$

Question:

My professor remarks that the last definition is independent of the choice of the sequence of simple functions $(u_n)$ with $u_n \uparrow f$. How can I verify this?


Here's what I tried:

Let $(u_n)$ and $(v_n)$ be two sequences of non-negative simple functions with $u_n \uparrow f$ and $v_n \uparrow f$.

Since $u_1 \leq u_2 \leq,\ldots$ and $v_1 \leq v_2 \leq,\ldots$, $\int u_n \, d\mu$ and $\int v_n \, d\mu$ are increasing sequences. Therefore the limits of both sequences exist (possibly infinite) and it suffices to show that $\forall m \in \mathbb{N} \, \forall \gamma \in (0,1)$:

$$\lim_{n\rightarrow \infty} \int u_n \, d\mu \geq \gamma \int v_m \, d\mu$$

Let $m \in \mathbb N, \gamma \in (0,1)$ and $x \in M$. Since $u_n(x)$ converges to $f(x)$ from below and $\gamma v_n(x)$ converges to $\gamma f(x)$ from below there exists $N \in \mathbb N$ such that $\forall n>N$: $$u_n(x) \geq \gamma v_m(x)$$

But now I'm stuck. I want to conclude that $\int u_n \, d\mu \geq \gamma \int v_m \, d\mu$, but I can't do that because my $n$ from the equation in previous line depends on the point $x$, so that I cannot be sure to find a finite $n$ such that $u_n(x) \geq \gamma v_m(x)$ holds for all $x \in M$!

Can my attempt be salvaged or is there a better strategy to prove my professor's remark?

Thank you!

$\endgroup$
  • 1
    $\begingroup$ Another definition is $\int f \, d\mu = \sup \{ \int u \, d\mu \mid u \text{ non-negative simple} \}$. Then there is no problem with welldefinedness. There might be other problems, though. $\endgroup$ – md2perpe Aug 28 '17 at 19:07
  • $\begingroup$ Kenny Wong also pointed this out, this definition appears to be the more common one. I think it should be $\sup \{\int u \, d\mu \, | \, u \text{ non-negative, simple and $u \leq f$} \}$, though, right? $\endgroup$ – Epiousios Aug 28 '17 at 20:25
  • 1
    $\begingroup$ You are of course right that I had missed $u \leq f$. $\endgroup$ – md2perpe Aug 28 '17 at 20:29
  • 1
    $\begingroup$ Strange, and I think awkward, definition. The usual one is$$\int f = \sup \int s$$ where the supremum is taken over all nonnegative simple functions $s\le f.$ This is a familiar idea given the Darboux approach to the Riemann integral. $\endgroup$ – zhw. Aug 28 '17 at 21:51
2
$\begingroup$

Yes, this type of argument can be salvaged!

Fix a value of $m \in \mathbb N$, and fix a value of $\gamma \in (0, 1)$.

Consider the measurable sets $$ E_n = \{ x : u_n(x) \geq \gamma v_m(x) \}.$$ These sets form a "nested family"; that is, they obey $$ E_1 \subset E_2 \subset E_3 \subset \dots$$

Furthermore, since $u_n \uparrow f$ and $v_m \leq f$, we also have $$ \bigcup_{n \in \mathbb N} E_n = M,$$ where $M$ is the entire measurable space.

Now for each $n \in \mathbb N$, we have $$ \int_M u_n \ d\mu \geq \int_{E_n} u_n \ d\mu \geq \gamma \int_{E_n}v_m \ d\mu.$$ Taking the limit $n \to \infty$ on both sides, we have $$ \lim_{n \to \infty} \int_M u_n d\mu \geq \gamma \lim_{n \to \infty} \int_{E_n} v_m \ d\mu \ \ \ \ \ \ \ \ (\ast).$$

Now suppose that the simple function $v_m$ can be written as $$ v_m = \sum_{i = 1}^p \alpha_i 1_{A_i},$$ where the $\alpha_i$ are constants and the $A_i$ are measurable sets. Then: $$ \int_{E_n} v_m \ d\mu = \sum_{i = 1}^p \alpha_i \mu(A_i \cap E_n).$$

However, for each $i$, we have $$ A_i \cap E_1 \subset A_i \cap E_2 \subset A_i \cap E_3 \subset \dots$$ and $$ \bigcup_{n \in \mathbb N} (A_i \cap E_n) = A_i.$$ Therefore, by countable additivity of the measure, it follows that $$ \lim_{n \to \infty} \mu(A_i \cap E_n) = \mu(A_i),$$ for each $i$, which implies that $$ \lim_{n \to \infty} \int_{E_n} v_m \ d\mu = \sum_{i = 1}^p \alpha_i \mu(A_i) = \int_{M} v_m \ d\mu.$$

Substituting this result into our earlier inequality $(\ast)$, we obtain $$ \lim_{n \to \infty} \int_{M} u_n \ d\mu \geq \gamma \int_M v_m \ d\mu.$$ Finally, we recall that $\gamma \in (0, 1)$ and $m \in \mathbb N$ were chosen arbitrarily, so we can conclude that $$ \lim_{n \to \infty} \int_M u_n \ d\mu \geq \lim_{m \to \infty} \int_M v_m \ d\mu.$$ We can get the reverse inequality by reversing the roles of $u_n$ and $v_m$, and this completes the proof.


By the way, the usual definition of the Lebesgue integral is that $ \int_M f \ d\mu $ is the supremum of $\int_M \varphi \ d\mu$, where the supremum is taken over all simple $\varphi$ such that $0 \leq \varphi \leq f$. With this definition, it is possible to prove something called the Monotone Convergence Theorem, which states that if $f_n $ is a positive sequence such that $f_n \uparrow f$, then $\lim_{n \to \infty} \int_M f_n = \lim_M f$. Applying this theorem to $u_n \uparrow f$ and also to $v_m \uparrow f$, you learn that $\lim_{n \to \infty} \int_M u_n \ d\mu$ and $\lim_{m \to \infty} \int_M v_m \ d\mu$ are both equal to my definition of $\int_M f \ d\mu$ (defined using the supremum), and hence, are equal to one another. In fact, my attempt at completing your argument steals lots of ideas from the proof of this more general Monotone Convergence Theorem.

$\endgroup$

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.