I have just read a textbook on stochastic processes that implicitly uses the fact that \begin{equation} \int \liminf_{t \to \infty} f_t \leq \liminf_{t \to \infty} \int f_t, \end{equation} for non-negative measurable functions $\{f_t\}_{t \geq 0}$. Most textbooks state this theorem (Fatou's lemma) in the discrete case.

Is it valid that the proof in the discrete case can be used in the continuous case?

Can monotone convergence theorem and dominated convergence theorem be generalised to the continuous case too?

  • $\begingroup$ what do you mean with discrete case and continuous case? Fatou's Lemma, MCT and DCT are proven for continuous functions. $\endgroup$ – Bman72 Oct 5 '14 at 11:44
  • 3
    $\begingroup$ Its true for every sequence $t_n \to \infty$ by the standard statement. But this means its true for the continuous limit too, since $f(t) \to a$ as $t \to c$ iff $f(t_n) \to a$ for every sequence $t_n \to a$. $\endgroup$ – user38355 Oct 5 '14 at 11:44
  • $\begingroup$ And yes, the same trick works for monotone convergence and dominated convergence. $\endgroup$ – user38355 Oct 5 '14 at 11:51

What @brom writes is essentially true:

Take some sequence $t_n \to \infty$ such that $\lim_n \int f_{t_n} d\mu \to \liminf_{t \to \infty} f_t d\mu$.

We then have $\liminf_{ t \to \infty} f_t \leq \liminf_n f_{t_n}$ and hence we have that

$$ \int \liminf_{t \to \infty} f_t d\mu \leq \int \liminf_n f_{t_n} \leq \liminf_n \int f_{t_n} d\mu = \liminf_{t \to \infty}\int f_t d\mu, $$

where we used the "ordinary" Fatou. The above is true if $\liminf_t f_t$ is measurable.

But to ensure this, we have to place additional assumptions. For example, let $V \subset \Bbb{R}$ be a non-measurable set. Define

$$ f_t (x) := -\chi_{V \times \Bbb{N}} (x,t). $$

Then $(x,t) \mapsto f_t (x)$ is Lebesgue-measurable (because $V \times \Bbb{N}$ is a null-set), but we easily see that

$$ \liminf_{t \to \infty} f_t(x) = -\chi_V (x), $$

which is not measurable. I am not completely certain wether one can construct a similar counterexample in which $(x,t) \mapsto f_t (x)$ is even Borel-measurable.

For the dominated convergence theorem, the situation is easier. Let us assume that $\lim_{t \to \infty} f_t(x) \to f(x)$ for almost every $x$ and where each $f_t$ is measurable and that $|f_t (x)| \leq g(x)$ with $g \in L^1$.

Then let $t_n \to \infty$ be any sequence converging to $\infty$. This implies that $f(x) = \lim_n f_{t_n}(x)$ is measurable and ordinary dominated convergence yields

$$ \lim_n \int f_{t_n} d\mu = \int f d\mu. $$

As this holds for arbitrary $t_n \to \infty$, we derive

$$ \lim_{t \to \infty} \int f_{t}d\mu = \int f d\mu. $$

For monotone convergence, the argument is essentially the same (again without the measurability problem).

EDIT: I have now figured out the behaviour for Borel-measurable functions. Note that the following uses some facts from descriptive set theory, in particular about analytic sets. Also, I only prove my claims in the setting of $\Bbb{R}^2$, but the proof should generalize easily to general polish spaces.

In general, $\liminf_{t\to\infty}f_{t}\left(x\right)$ will not be Borel measurable, even if $\left(x,t\right)\mapsto f_{t}\left(x\right)$ is. To see this, it suffices to show the claim for $\limsup$ instead of $\liminf$. To this end, let $A\subset\left[0,\frac{1}{2}\right]^{2}$ be a Borel-set for which the projection $\pi_{1}\left(A\right)\subset\left[0,\frac{1}{2}\right]$ is not a Borel-set. Here, $\pi_{1}\left(x,y\right):=x$. The existence of such a set is a well-known fact in descriptive set theory. Then $$ B:=\bigcup_{n\in\mathbb{N}}A+\left(0,n\right)\subset\left[0,\frac{1}{2}\right]\times\mathbb{R} $$ is a Borel-set. Define $f_{t}\left(x\right):=\chi_{B}\left(x,t\right)$. For $x\in\left[0,\frac{1}{2}\right]$, there are now two possibilities:

  1. We have $x\in\pi_{1}\left(A\right)$, i.e. there is some $y\in\left[0,\frac{1}{2}\right]$ such that $\left(x,y\right)\in A$. This implies $\left(x,y+n\right)\in B$ for all $n\in\mathbb{N}$ and hence $$ \limsup_{t\to\infty}f_{t}\left(x\right)\geq\lim_{n}f_{y+n}\left(x\right)=1. $$

  2. We have $x\notin\pi_{1}\left(A\right)$, i.e. $\left(x,y\right)\notin A$ for all $y\in\mathbb{R}$. If we have $f_{t}\left(x\right)>0$ for some $t\in\mathbb{R}$, this would imply $\left(x,t\right)\in A+\left(0,n\right)$ for some $n$ and hence $\left(x,t-n\right)\in A$, a contradiction. Hence, $$ \limsup_{t\to\infty}f_{t}\left(x\right)\leq0. $$

Now define $g\left(x\right):=\limsup_{t\to\infty}f_{t}\left(x\right)$ for $x\in\mathbb{R}$. The above implies that $$ g^{-1}\left(\left(\frac{1}{2},\infty\right)\right)=\pi_{1}\left(A\right) $$ is not Borel-measurable.

But $g=\limsup_{t\to\infty}f_{t}$ will be Lebesgue-measurable. To see this, first note $$ g\left(x\right)=\lim_{n\to\infty}\sup_{t\geq n}f_{t}\left(x\right), $$ so that it suffices to show that $g_{n}:=\sup_{t\geq n}f_{t}$ is Lebesgue-measurable for each $n\in\mathbb{N}$. To see this, let $\alpha\in\mathbb{R}$ be arbitrary and $f\left(x,t\right):=f_{t}\left(x\right)$. We have \begin{eqnarray*} g_{n}^{-1}\left(\left(\alpha,\infty\right]\right) & = & \left\{ x\in\mathbb{R}\,\mid\,\exists t\geq n:\: f_{t}\left(x\right)>\alpha\right\} \\ & = & \left\{ x\in\mathbb{R}\,\mid\,\exists t\in\mathbb{R}:\:\left(x,t\right)\in f^{-1}\left(\left(\alpha,\infty\right]\right)\cap\left(\mathbb{R}\times\left[n,\infty\right)\right)\right\} \\ & = & \pi_{1}\left(f^{-1}\left(\left(\alpha,\infty\right]\right)\cap\left(\mathbb{R}\times\left[n,\infty\right)\right)\right). \end{eqnarray*} But $f$ is Borel-measurable by assumption, so that the set above is an analytic set. But from descriptive set theory, we know that all analytic sets are Lebesgue measurable, so that each $g_{n}$ is Lebesgue measurable


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