Pointwise convergence and L1 convergence in bounded mass case

Question

I have a question regarding convergence modes and their relationships, my problem is actually an application to probability, How to prove that :

${f_n}$ and $g$ are probability density functions such that for all $x \in R$ ${f_n}(x) \rightarrow g(x)$ as $n \rightarrow \infty$, then ${f_n}$ converges to $g$ in $L^{1}$

I don't see how to prove this, my first thought was to use the bounded convergence theorem but I don't see how (similarly with Fatou's lemma), I think it is more subtle. I found this theorem however which is more general, but I can't find its proof :

Suppose that ${f_n: X \rightarrow [0,+\infty)}$ are measurable, are such that ${\sup_n \int_X f_n\ d\mu < \infty}$, and converge pointwise almost everywhere to some measurable limit ${f: X \rightarrow [0,+\infty)}$. Show that ${f_n}$ converges in ${L^1}$ norm to ${f}$ if and only if ${\int_X f_n\ d\mu}$ converges to ${\int_X f\ d\mu}$. Informally, we see that in the unsigned, bounded mass case, pointwise convergence implies ${L^1}$ norm convergence if and only if there is no loss of mass.

Do you see how ?

Thanks

Answer 1

Let $f_n: X \to [0, \infty)$ be measurable functions s.t. $\operatorname{sup}_n \int_X f_n d \mu < \infty$ and $f_n \to f$ a.e. where $f: X \to [0, \infty)$ measurable. I'll use the notation $\|f\|_1$ for $\int_X |f|$.

First, assume that $f_n \to f$ in $L^1$ norm. Then $$ \|f_n\|_1 = \|f_n - f + f\|_1 \leq \|f_n - f\|_1 + \|f\|_1\,. $$ Now, because $f_m \to f$ in the $L^1$ norm, $\|f_n-f\|_1 \to 0$ as $n \to \infty$. Hence $\|f_n\|_1 \to \|f\|_1$.

Now, assume that $\|f_n\|_1 \to \|f\|_1$. We want to show that $\lim_n \|f_n - f\|_1 = 0$.

First recall Fatou's lemma: if $(g_n)$ is a sequence of non-negative measurable functions, then $$ \int_X \liminf_n g_n \leq \liminf_n \int_X g_n\,. $$ We want to find a relation with this and the limit $\lim \|f_n-f\|_1$. Note that the following equivalence chain holds: \begin{align*} \liminf_n \left( - \int_X |f_n-f| \right) = 0 & \iff - \limsup_n \int_X |f_n - f| = 0\\ & \iff \limsup_n \int_X |f_n-f| = 0 \\ & \iff \lim_n \int_X |f_n-f| = 0\,. \end{align*} Now we found a link between Fatou's lemma and our limit: we just need to find a sequence $g_n$ which is of the form $$ g_n = a_n - |f_n-f| $$ and which satisfies the assumptions of Fatou's lemma. So we need $g_n \geq 0$. Let's try to achieve this by finding an upper bound for $|f_n - f|$. That's easy: $|f_n - f| \leq |f_n| + |f|$. Now set $a_n = |f_n| + |f|$. Now the sequence $(a_n)$ has the property $a_n \to a := 2 |f|$ and $\int a_n \to \int a$, which turns out to be crucial!

Let's plug our $g_n$ into Fatou: \begin{align*} \int_X \liminf_n (|f_n| + |f| - |f_n - f|) & \leq \liminf_n \int_X (|f_n| + |f| - |f_n-f|) \\ \implies \int_X \left( \liminf_n |f_n| + |f| + \liminf_n (-|f_n-f|) \right) &\leq \liminf_n \int_x |f_n| + \int_X |f| - \limsup_n \int_X |f_n-f| \\ \implies 2\int_X |f| & \leq 2 \int_X |f| - \limsup_n \int_X |f_n-f| \\ \implies \limsup_n \int_X |f_n-f| & \leq 0\,. \end{align*} (Note that if $b_n \to b$, then $\liminf_n (b_n + c_n) = b + \liminf_n c_n$.) Because $\lim_n |f_n-f| \leq \limsup_n |f_n-f|$, we have $\lim_n |f_n-f| = 0$, so $f_n \to f$ in $L^1$ norm.

P.S. this proof is written in a form that is easy to generalize to the $L^p$ case.

Answer 2

The theorem you found is known as Scheffé's lemma. Here's a nice proof.

If $f_n$ and $g$ are probability density functions then they all integrate to 1, so you get $\int f_n\to\int f$ automatically.