Let $f_1, f_2, \cdots$ and $f$ be nonnegative lebesgue integrable functions on $\mathbb{R}$ such that $$\lim_{n \to \infty}\int_{-\infty}^y f_n(x)dx = \int_{-\infty}^y f(x)dx \; \; \text{ for each $y \in \mathbb{R}$}$$ and $$\lim_{n \to \infty}\int_{-\infty}^{\infty} f_n(x)dx = \int_{-\infty}^{\infty} f(x)dx $$

Then I want to prove that $ \liminf_{n \to \infty} \int_{U}f_n(x)dx \geq \int_{U}f(x)dx\:$ for any open set $U$ of $\mathbb{R}.$

I think this can done by proving $\lim_{n \to \infty}f_n(x) = f(x) $ and then using Fatou's lemma. But I am not able to prove $\lim_{n \to \infty}f_n(x) = f(x) $ .

Thank you in advance.

  • $\begingroup$ In fact, if you had $f_n \rightarrow f$ pointwise, then you get the much stronger statement that $\int_A f_n \rightarrow \int_A f$ for every measurable set $A$. $\endgroup$ – Rick Sanchez Apr 28 '16 at 5:50
  • 1
    $\begingroup$ In fact, it may be false that $\lim_{n \to \infty}f_n(x) = f(x) $. $\endgroup$ – Ramiro Apr 29 '16 at 11:27
  • $\begingroup$ [But I am not able to prove $\lim_{n \to \infty}f_n(x) = f(x) $]---> in fact, as said by Ramiro, this may be false because $f$ is determined up to addition with a negligible function (as the characteristic function of $\mathbb{Q}$). $\endgroup$ – Duchamp Gérard H. E. Dec 15 '16 at 15:42
  • 1
    $\begingroup$ Idea: prove it for intervals and then use that open sets are countable disjoint unions of intervals. $\endgroup$ – Jose27 Dec 16 '16 at 8:32

[But I am not able to prove $\lim\limits_{n\to \infty} f_n(x)=f(x)$], as said by Ramiro and Sir Gerard, it may be false. For example

Suppose that $g(x)$ is a integrable function which is defined on $[1,2]$. We can denote that $f(x) = g(x)$ if $x \in [1,2]$, else $f(x) = 0$. Then $\int\limits_U g(x) dx = \int\limits_U f(x)$ for any $U \subset \mathbb{R}$. Now, for any $k \in \mathbb{N}_+$, take

$f_k(x) = 1 $ if $x \in [0, 1/k]$, else, $f_k(x) = f(x)$. Then $\lim\limits_{k \to + \infty} \int\limits_{-\infty}^{y} f_k(x)dx = \int\limits_{-\infty}^y f(x) dx$ for any $y \in \mathbb{R} \cup \{+ \infty\}$. But $\lim\limits_{k \to + \infty} f_k(x) \neq f(x)$ because $\lim\limits_{k \to +\infty} f_k(x) =1$ at $x = 0$.

  • $\begingroup$ This is not a particularly good example, as your sequence still converges almost surely. $\endgroup$ – Dominik Dec 16 '16 at 8:34

This is a particular instance of the Portmanteau theorem.

Proof of the statement. Let $F_n(x) = \int_{-\infty}^{x} f_n(t) \, dt$ and $F(x) = \int_{-\infty}^{x} f(t) \, dt$. Then for all test functions $\varphi \in C_c^{\infty}(\Bbb{R})$, integration by parts and the bounded convergence theorem shows that

\begin{align*} \int_{\Bbb{R}} \varphi(x) f_n(x) \, dx &= -\int_{\Bbb{R}} \varphi'(x) F_n(x) \, dx \\ &\xrightarrow[n\to\infty]{} -\int_{\Bbb{R}} \varphi'(x) F(x) \, dx = \int_{\Bbb{R}} \varphi(x) f(x) \, dx \end{align*}

Now let $U$ be open. Then there exists a sequence of test functions $(\varphi_n) \subset C_c^{\infty}(\Bbb{R})$ such that $0 \leq \varphi_n \leq 1$ and $\varphi_n \uparrow \mathbf{1}_U$ as $n\to\infty$. Thus for any fixed $m$, we have

\begin{align*} \int_{\Bbb{R}} \varphi_m (x) f(x) \, dx &= \liminf_{n\to\infty} \int_{\Bbb{R}} \varphi_m (x) f_n(x) \, dx \\ &\leq \liminf_{n\to\infty} \int_{\Bbb{R}} \mathbf{1}_U(x) f_n(x) \, dx \\ &= \liminf_{n\to\infty} \int_{U} f_n(x) \, dx. \end{align*}

Now taking $m \to \infty$ and utilizing the monotone convergence theorem shows the claim. ////

Next, let us discuss a counterexample of pointwise convergence. Let

$$ f_n(x) = \begin{cases} 1 + \cos(2n\pi x),& 0 \leq x \leq 1 \\ 0,& \text{otherwise} \end{cases}, \qquad f(x) = \mathbf{1}_{[0,1]}(x) $$

and define $(F_n)$ and $F$ as before. Then it is easy to check that $F_n$ converges to $F$ pointwise, but $f_n(x)$ does not converge if $x \in [0, 1] \setminus \Bbb{Q}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.