Monotone Convergence Theorem (Measure Theory)/Corollary

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Corollary to Monotone Convergence Theorem (Measure Theory)

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $u : X \to \overline \R_{\ge 0}$ be a positive $\Sigma$-measurable function.

Let $\sequence {u_n}_{n \mathop \in \N}$ be an sequence of positive $\Sigma$-measurable functions $u_n : X \to \overline \R_{\ge 0}$ such that:

$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

for $\mu$-almost all $x \in X$.

Suppose also that $u_1$ is $\mu$-integrable function.


Then $u_n$ is $\mu$-integrable for each $n \in \N$ and $u$ is $\mu$-integrable with:

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$


Proof

Suppose that:

$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

for all $x \in X$.

Then in particular we have:

$u_1 \ge u_n$ for each $n \in \N$.

From Integral of Positive Measurable Function is Monotone, we have:

$\ds \int u_1 \rd \mu \ge \int u_n \rd \mu$

Since $u_1$ is $\mu$-integrable, we have that:

$u_n$ is $\mu$-integrable for each $n \in \N$.

We also have:

$u_1 \ge u$

so $u$ is $\mu$-integrable.

From Integrable Function is A.E. Real-Valued, for each $k \in \N$ there exists a $\mu$-null set $N_k \subseteq X$ such that:

whenever $\map {u_k} x = \infty$, we have $x \in N_k$.

Let:

$\ds N = \bigcup_{k \mathop = 1}^\infty N_k$

then:

$\map {u_k} x < \infty$ for each $x \in X \setminus N$ for all $k \in \N$.

Since $\sigma$-algebras are closed under countable union, we have:

$N$ is $\Sigma$-measurable.

From Null Sets Closed under Countable Union, we have:

$N$ is $\mu$-null.


Now let $\sequence {v_n}_{n \mathop \in \N}$ be a sequence of real-valued functions $v_k : X \to \R$ with:

$\map {v_k} x = \map {u_1} x \times \map {\chi_{X \setminus N} } x - \map {u_k} x \times \map {\chi_{X \setminus N} } x$

We show that $\sequence {v_n}_{n \mathop \in \N}$ is an increasing sequence converging to:

$v = u_1 \times \chi_{X \setminus N} - u \times \chi_{X \setminus N}$

which we will verify is well-defined.

For $x \in X \setminus N$, we then have:

$\map {v_k} x = \map {u_1} x - \map {u_k} x$

Since:

$\sequence {\map {u_k} x}_{k \mathop \in \N}$ is an decreasing sequence

we have:

$\sequence {\map {v_k} x}_{k \mathop \in \N}$ is a increasing sequence.

From Monotone Convergence Theorem (Real Analysis): Decreasing Sequence, we have:

$\map u x < \infty$

So we have:

$\ds \lim_{k \mathop \to \infty} \map {v_k} x = \map {u_1} x - \map u x = \map {u_1} x \times \map {\chi_{X \setminus N} } x - \map u x \times \map {\chi_{X \setminus N} } x$

for each $x \in X \setminus N$.

Now suppose that $x \in N$.

Then we have:

$\map {v_k} x = 0$

for each $k \in \N$.

Clearly, $\sequence {\map {v_k} x}_{k \mathop \in \N}$ is increasing and converges to:

$0 = \map {u_1} x \times \map {\chi_{X \setminus N} } x - \map u x \times \map {\chi_{X \setminus N} } x$

So we have shown that $\sequence {v_n}_{n \mathop \in \N}$ is an increasing sequence with:

$v_n \to u_1 \times \chi_{X \setminus N} - u \times \chi_{X \setminus N}$

By the Monotone Convergence Theorem, we therefore have:

$\ds \int \paren {u_1 \times \chi_{X \setminus N} - u \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \int v_n \rd \mu$

That is:

$\ds \int \paren {u_1 \times \chi_{X \setminus N} - u \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \paren {u_1 \times \chi_{X \setminus N} - u_n \times \chi_{X \setminus N} } \rd \mu$

From Integral of Integrable Function is Additive: Corollary 2, we have:

$\ds \int \paren {u_1 \times \chi_{X \setminus N} } \rd \mu - \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \int \paren {u_1 \times \chi_{X \setminus N} } \rd \mu - \lim_{n \mathop \to \infty} \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu$

giving:

$\ds \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu$

From Characteristic Function of Null Set is A.E. Equal to Zero: Corollary, we have:

$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.

So, from Pointwise Multiplication preserves A.E. Equality, we have:

$u \times \chi_{X \setminus N} = u$ $\mu$-almost everywhere

and:

$u_n \times \chi_{X \setminus N} = u_n$ $\mu$-almost everywhere for each $n \in \N$.

From A.E. Equal Positive Measurable Functions have Equal Integrals, we have:

$\ds \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \int u \rd \mu$

and:

$\ds \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu = \int u_n \rd \mu$ for each $n \in \N$.

So we obtain:

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$


Now suppose that:

$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

for $\mu$-almost all $x \in X$.

Then, there exists a $\mu$-null set $N_1 \subseteq X$ such that whenever:

$\map {u_i} x < \map {u_j} x$ for some $i < j$

and:

either $\ds \lim_{n \mathop \to \infty} \map {u_n} x$ does not exist or $\ds \map u x \ne \lim_{n \mathop \to \infty} \map {u_n} x$

we have $x \in N$.

For each $n \in \N$, define $v_n : X \to \overline \R_{\ge 0}$ by:

$\map {v_n} x = \map {u_n} x \times \map {\chi_{X \setminus N_1} } x$

for each $x \in X$.

Also define $v : X \to \overline \R_{\ge 0}$ by:

$\map v x = \map u x \times \map {\chi_{X \setminus N_1} } x$

for each $x \in X$.

Clearly, if $x \in X \setminus N$, we have:

$\map {v_n} x = \map {u_n} x$ for each $n$

and:

$\map v x = \map u x$

From the definition of $N$, we have:

$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

for $x \in X \setminus N$.

So:

$\map {v_i} x \ge \map {v_j} x$ for all $i \le j$

and:

$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$

If $x \in N$, we have:

$\map {v_n} x = 0$ for each $n \in \N$

and:

$\map v x = 0$

So we have:

$\map {v_i} x \ge \map {v_j} x$ for $i \le j$

and:

$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$

for $x \in N$.

So, we have:

$\map {v_i} x \ge \map {v_j} x$ for $i \le j$

and:

$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$

for all $x \in X$.

From our previous work, we have:

$\ds \int v \rd \mu = \lim_{n \mathop \to \infty} \int v_n \rd \mu$

That is:

$\ds \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu$

From Characteristic Function of Null Set is A.E. Equal to Zero: Corollary, we have:

$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.

So, from Pointwise Multiplication preserves A.E. Equality, we have:

$u \times \chi_{X \setminus N} = u$ $\mu$-almost everywhere

and:

$u_n \times \chi_{X \setminus N} = u_n$ $\mu$-almost everywhere for each $n \in \N$.

Then from A.E. Equal Positive Measurable Functions have Equal Integrals, we have:

$\ds \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \int u \rd \mu$

and:

$\ds \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu = \int u_n \rd \mu$ for each $n \in \N$.

So, we get:

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$

$\blacksquare$


Sources

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