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 $A$. $\endgroup$ Apr 28, 2016 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, 2016 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 $\mathbb{Q}$). $\endgroup$ Dec 15, 2016 at 15:42
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    $\begingroup$ Idea: prove it for intervals and then use that open sets are countable disjoint unions of intervals. $\endgroup$
    – Jose27
    Dec 16, 2016 at 8:32

2 Answers 2


[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, 2016 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 $x \in [0, 1] \setminus \Bbb{Q}$.


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