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Theorem $\mathbf{A.2.11}$ (Dominated convergence). Let $f_n : X \to \mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X \to \mathbb R$ such that $|f_n(x)| \leq |g(x)|$ for $\mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $\mu$-almost every point to some function $f : X \to \mathbb R$. Then $f$ is integrable and satisfies $$\lim_n \int f_n \, d\mu = \int f \, d\mu.$$

I wanted to know if in the hypothesis $|f_n(x)| \leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?

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    $\begingroup$ The dominating function $g$ is the whole point of the theorem. It's in the very name of the theorem. $\endgroup$ Mar 31, 2019 at 3:30

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This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence $$ f_n(x) := \frac{1}{n} \mathbf{1}_{[0,n]}(x). $$ Clearly, $f_n \in L^1(\mathbb{R})$ for each $n \in \mathbb{N}$. Moreover, $f_n(x) \to 0$ as $n \to \infty$ for each $x \in \mathbb{R}$. However, \begin{align*} \lim_{n \to \infty} \int_{\mathbb{R}} f_n\,\mathrm{d}m = \lim_{n \to \infty} \int_0^n \frac{1}{n}\,\mathrm{d}x = 1 \neq 0. \end{align*}

Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < \infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.

Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,\mathfrak{M},\mu)$ converging almost everywhere to a measurable function $f$. If $E \in \mathfrak{M}$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < \infty$, then $$ \lim_{n \to \infty} \int_E f_n\,\mathrm{d}\mu = \int_E f\,\mathrm{d}\mu. $$ In fact, one has $f_n \to f$ strongly in $L^1(E)$.

In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.

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  • $\begingroup$ I understood. Thanks a lot for the help $\endgroup$
    – Ilovemath
    Mar 31, 2019 at 1:45
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In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = \chi_{[n,n+1]}$ on $\mathbf R_{\ge 0}$ are all integrable, and $f_n(x) \to 0$ for all $x\in \mathbf R_{\ge 0}$, but they are not dominated by an integrable function $g$, and indeed we do not have $$ \lim_{n\to\infty} \int f_n = \int \lim_{n\to\infty}f_n $$ since in this case, the left-hand side is $1$, but the right-hand side is $0$.


To see why there is no dominating function $g$, such a function would have the property that $g(x)\ge 1$ for each $x\ge 0$, so it would not be integrable on $\mathbf R_{\ge 0}$.

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  • $\begingroup$ I understood. Thanks a lot for the help $\endgroup$
    – Ilovemath
    Mar 31, 2019 at 1:46
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As you guessed, you actually don't need a dominating function, but you need some condition in addition to the integrability and convergence of $f_{n}$.

Here's a simple example that shows that you don't need a dominating function. Let $X=(0,1]$ and $\mu$ be the Lebesgue measure. Let $$ f_{n}(x) = \begin{cases} [1/n - 1/(n+1)]/n & \text{if $x \in (1/(n+1),1/n]$,} \\ 0 & \text{otherwise.} \end{cases} $$ Then $\int f_{n} = 1/n$ and $\lim_{n} f_{n} = 0$. In this case, there's no dominating function that is integrable, but $$ \lim_{n} \int f_{n} = \int \lim_{n} f_{n} = 0. $$

There're various extensions of the dominating convergence theorem, and there's a recent paper that gives a necessary and sufficient condition for the first equality above to be valid. The exact condition is somewhat complicated, but it's strictly weaker than uniform integrability, which is also somewhat complicated and in turn weaker than the hypothesis of the dominated convergence theorem.

(This answer is almost identical to another answer of mine, but I'm a newbie here and I've just realized that a post that gives a link that answers the question can be deleted.)

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