Lebesgue Integration Part 3- Integration of Nonnegative Functions

Let’s first define the integral of a nonnegative simple function given by s=\sum^{n}_{i=1}\alpha_i\chi_{A_i}, where s takes on the values \alpha_1,...,\alpha_n on sets A_1,...,A_n. Then the integral of s over a subset E in the \sigma-algebra is given by the geometrically intuitive \int_E s \ d\mu = \sum^n_{i=1}\alpha_i\mu(A_i\cap E). We can then extend this to arbitrary nonnegative measurable functions f to get \int_E f \ d\mu = \sup\int_E s \ d\mu, where the supremum is taken over all simple functions s such that 0\le s\le f. We can easily deduce familiar properties that if 0\le f\le g over E, the integral of g over E is greater than that of f, that the the Lebesgue integral is increasing by set inclusion and multiplicative by scalar factors, and that \int_E f \ d\mu = 0 if E is of measure 0 or f is identically zero over E.

Recall from the previous post that to every nonnegative function f corresponds some increasing sequence of simple nonnegative functions that converges to f. This result and the following result empower us to prove many results on the integral of arbitrary nonnegative functions simply by proving them for the integral of simple functions.

Result 8 (Monotone Convergence): If \{f_n\} is an increasing sequence of nonnegative, measurable functions, then

f_n\to f\Rightarrow \int_X f_n \ d\mu\to\int_X f \ d\mu.

Proof: By a previous result, f is certainly measurable. Let’s say that the sequence of integrals converges to some L. We know that L\le\int_X f \ d\mu and thus want to prove that this inequality is true in the reverse. First observe that the function \phi(E)=\in_E s \ d\mu is a measure for s simple (we leave it as an exercise to the reader to verify that \phi is countably additive and that \phi(\emptyset)=0). We will proceed to construct an increasing sequence of sets whose union is X and use this to show that \alpha is greater than the integral of any simply function bounded between 0 and f. Fix a constant 0<c<1 and any simple measurable function 0\le s\le f, and define E_n = \{x: f_n(x)\ge cs(x). We can verify that \{E_n\} is an increasing sequence of measurable sets whose union is the whole space. So if we take the limit of both sides of \int_X f_n \ d\mu\ge c\int_{E_n} s \ d\mu, we get that \alpha\ge c\int_X s \ d\mu. This holds for all 0<c<1 and all simple 0\le s\le f, so we have equality. Note that if we had picked c=1, the union of the sets E_n would not necessarily have been the entire space (pick f=s), so the strict inequality of 0<c<1 was necessary.

Let’s see an instance of monotone convergence in action. It is straightforward to prove that the Lebesgue integral is countably additive for simple functions, but we can extend this to arbitrary functions:

Result 9: \int_X\left(\sum^{\infty}_{n=1}f_n\right) \ d\mu = \sum^{\infty}_{n=1}\int_X f_n \ d\mu.

Proof: It suffices to prove that \int_X (f+g) \ d\mu = \int_X f \ d\mu + \int_X g \ d\mu, after which we are done by induction. Take an increasing sequence \{s_n\} of simple measurable functions that converge to f and one that converges to g. Clearly the sequence \{s_n+t_n\} is an increasing sequence of measurable functions that converges to f+g, so by monotone convergence, we are done.

As a corollary, not only is the integral of a nonnegative, simple function a measure, but the integral of any nonnegative function is a measure as well. What we get is a cool equality calling to mind a change-of-variables formula.

Result 10: If \phi(E) = \int_E f \ d\mu, then \int_E g \ d\mu = \int_E gf \ d\mu for all g measurable and nonnegative. Note: gf denotes multiplication, not composition.

Proof: We will prove this for simple g and then use monotone convergence to prove for all g. Again, say that g takes on values \alpha_1,...,\alpha_n over regions E_1,...,E_n. Indeed,

\displaystyle\int_E g \ d\mu = \sum^{\infty}_{i=1}\alpha_i\phi(E_i) = \sum^{\infty}_{i=1}\alpha_i\displaystyle\int_{E_i} f \ d\mu = = \displaystyle\int_E \sum^{\infty}_{i=1} \chi_{E_i}\alpha_if = \displaystyle\int_E gf \ d\mu.

The next result establishes that the integral of the upper limit of a sequence of functions is at most the upper limit of the integrals of those functions.

Result 11 (Fatou’s Lemma): \int_X\left(\liminf_{n\to\infty} f_n\right) \ d\mu \le \liminf_{n\to\infty}\int_X f_n \ d\mu.

Remark: At first I had trouble remembering the direction of the inequality, but I chanced upon this! Of course, the best way to “carve this into your soul” would be to use it frequently, but that will come with time. 🙂 Anyways, on to the proof…

Proof: This result just screams “monotone convergence,” so define g_k = \inf_{n\ge k}f_n. By definition \{g_n\} is an increasing sequence of nonnegative, measurable functions that converges to \liminf_{n\to\infty}f_n, so by monotone convergence, \int_X\liminf_{n\to\infty} f_n \ d\mu = \lim_{n\to\infty}\int_X g_n \ d\mu. It suffices to prove that the right side is no greater than \liminf_{n\to\infty}\int_X f_n \ d\mu. Yet by definition, \int_X g_k \ d\mu\le \int_X f_n \ d\mu for all n\ge k, so \int_X g_k \ d\mu\le \inf_{n\ge k}\int_X f_n \ d\mu. Taking the limit on both sides gives what we need.

With this machinery, we are ready to define integration of complex functions with little difficulty. The main result in my next post will be a cool variant on monotone convergence that makes use of Fatou’s Lemma.


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