The monotone convergence theorem, dominated convergence theorem, and Fatou’s Lemma are the three most important theorems in Lebesgue integration theory. In this article, we are going to learn the theorem and proof of the dominated convergence theorem and an example in detail.

Lebesgue’s Dominated Convergence Theorem

Consider (f_n(x))^∞_n=1 denotes a sequence of Lebesgue integrable functions that almost everywhere on I converge to a limit function f. Assume that g is a Lebesgue integrable function that ∣f_n∣ ≤ g nearly everywhere on I and for all n ∈ N. If lim_n→∞∫_If_n(x) dx = ∫_If(x)dx., then f is Lebesgue integrable on I.

It’s important to remember that lim_n→∞∫_I f_n(x) dx = ∫_I f(x) dx can alternatively be written as lim_n→∞∫_I f_n (x) dx = ∫_I lim_n→∞fn (x) dx.

Dominated Convergence Theorem Proof

Assume that (fn(x))n=1 is a sequence of Lebesgue integrable functions that nearly everywhere on I converge to a limit function f. Then, practically everywhere on I, the following equality holds:

Lim sup_n→∞f_n(x) = f(x) = lim inf_n→∞f_n(x) …(1)

Establish two new sequences of functions (G_n(x))^∞_n=1and (g_n(x))^∞_n=1with the following general terms:

G_n(x) = sup_k≥n{f_k(x)} and g_n(x) = inf_k≥n{f_k(x)} …(2)

Therefore (G_n(x))^∞_n=1 is a decreasing sequence of functions that almost always converges to f(x) on I. Furthermore, (g_n(x))^∞_n=1 is an expanding sequence of functions that almost always converges to f(x) on I. (G_n(x))∞_n=1and (g_n(x))^∞_n=1 are Lebesgue integrable function sequences.

As a result, almost everywhere on I, the following inequality holds:

g_n(x) ≤ f(x) ≤ G_n(x) …(3)

As ∣f_n(x)∣ ≤ g(x) exists practically everywhere on I, for any n∈N, and then for some Lebesgue integrable function g. As a result, nearly everywhere on I, the following chain of inequalities remains true:

−g(x) ≤ g_n(x) ≤ f_n(x) ≤ G_n(x) ≤ g(x) …(4)

Next, consider the function sequence (g−G_n)^∞_n=1. Then g−G_n is Lebesgue integrable on I and, moreover, is nearly everywhere on I an expanding sequence. In addition, the sequence (∫_I[g(x) − G_n(x)] dx)^∞_n=1 converges because:

∫_I[g(x) − G_n(x)] dx ≤ ∫_I2g(x) dx = 2 ∫_I g(x) dx < ∞ …(5)

We can see that the function g(x)−f(x) = lim_n→∞[g(x) − Gn(x)] is Lebesgue integrable on I using Levi’s Monotonic Convergence theorems concerning Lebesgue integrable functions, and that:

∫_I[g(x) − f(x)] = lim_n→∞∫_I[g(x) − G_n(x)]dx = ∫_Ig(x)dx − lim_n→∞∫_IG_n(x)dx = ∫_Ig(x)dx − ∫_If(x)dx …(6)

As a result, ∫_If(x)dx = lim_n→∞∫_IG_n(x)dx, and f is Lebesgue integrable on I.

Let’s consider the function sequence (g_n(x) + g(x))^∞_n=1. This is a sequence of Lebesgue integrable functions on I in increasing order. Moreover, ( ∫_I[g_n(x) + g(x)]dx)^∞_n=1, the sequence converges, because

∫_I[g_n(x) + g(x)] dx ≤ ∫_I2g(x)dx = 2∫_Ig(x)dx < ∞…(7)

So we know that g(x) + f(x) = lim_n→∞[g(x) + g_n(x)] is a Lebesgue integrable function according to Levi’s Convergence Theorem regarding Lebesgue integrable functions, that:

∫_I[g(x) + f(x)] dx = lim_n→∞∫_I[g(x) + g_n(x)] dx = ∫_Ig(x)dx + lim_n→∞∫_Ig_n(x)dx = ∫_Ig(x)dx + ∫_If(x)dx … (8)

This proves that f is Lebesgue integrable on I, and that ∫_If(x)dx = lim_n→∞∫_Ig_n(x) dx once more.

Because g_n(x) ≤ f(x) ≤ G_n(x) holds almost everywhere on I, the preceding chain of inequalities holds nearly everywhere on I:

∫_Ig_n(x)dx ≤ ∫_If(x)dx ≤ ∫_IG_n(x)dx lim_n→∞∫_Ig_n(x)dx ≤ lim_n→∞∫_If(x)dx ≤ lim_n→∞∫_IG_n(x)dx ∫_If(x)dx ≤ lim_n→∞∫_If(x)dx ≤ ∫_If(x)dx …(9)

Therefore, lim_n→∞∫_If_n(x)dx = ∫_If(x)dx is proved.

Aslo, read:

Solved Example

Example :

Prove that, for any n ≥ 2, the function,

\(\begin{array}{l}\frac{1}{(1+\frac{x}{n})^{n}x^{1/n}}\end{array} \)

is integrable on the interval [1, ∞), and calculate with justification.

Solution:

For n ∈ Z_>0and x ∈ [1, ∞), we can define,

\(\begin{array}{l}f_{n}(x)=\frac{1}{(1+\frac{x}{n})^{n}x^{1/n}}\end{array} \)

If for n ≥ 2 we have n − 1 ≥ n/2, and hence

\(\begin{array}{l}\left ( 1+\frac{x}{n} \right )^{n}\geq 1+n\frac{x}{n}+\left ( \frac{n(n-1)}{2} \right )\left ( \frac{x}{n} \right )^{2}\geq 1+x+\left ( \frac{n^{2}}{4} \right )\left ( \frac{x}{n} \right )^{2}= 1+x+\frac{x^{2}}{4}> \frac{x^{2}}{4}\end{array} \)

Hence,

\(\begin{array}{l}0\leq f_{n}x\leq \frac{1}{(1+\frac{x}{n})^{n}}\leq \frac{4}{x^{2}}\end{array} \)

Here, the function g(x) = 4/x² is integrable on the interval [1, ∞), and hence f_n is integrable on [0, ∞). As lim_n→∞(1 + x/n)ⁿ = e^x and lim_n→∞ x^1/n = 1 for all x ∈ [1, ∞), we can use DCT (Dominated Convergence Theorem) with the dominating function g to obtain the initial step in the preceding calculation:

\(\begin{array}{l}\displaystyle \lim_{ n\to \infty }\int_{1}^{\infty }\frac{1}{(1+\frac{x}{n})^{n}x^{1/n}}dx =\int_{1}^{\infty }e^{-x}dx =\frac{1}{e}\end{array} \)

And, the problem is now solved.

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Frequently Asked Questions on Dominated convergence Theorem

What does dominated convergence theorem mean?

The dominated convergence theorem states that “g” is a Lebesgue integrable function that ∣f_n∣ ≤ g nearly everywhere on I and for all n ∈ N. If lim_n→∞∫_If_n(x) dx = ∫_If(x)dx., then f is Lebesgue integrable on I.

Give the applications of the dominated convergence theorem.

It is widely utilized in probability theory, since it provides a necessary condition for the convergence of predicted values of random variables, in addition to its frequent presence in partial differential equations and mathematical analysis.

What does DCT mean in measure theory?

In measure theory, DCT means Dominated Convergence Theorem.