Based on Williams' Probability w/ Martingales:

Let $(S, \Sigma, \mu)$ be a measure space.

Scheffe's Lemma Part (ii): Suppose $\{f_n\}_{n \in \mathbb{N}}, f \in \mathscr{L}^1 (S, \Sigma, \mu)$ and $\lim_{n \to \infty} f_n(s) = f(s) \forall s \in S$ or a.e. in S. Then $$\lim_{n \to \infty} \int_S |f_n - f| d\mu = 0 \iff \lim_{n \to \infty} \int_S |f_n| d\mu = \int_S |f| d\mu$$

In proving Scheffe's Lemma, we could use Fatou's Lemmas to show that

$$\lim_{n \to \infty} \int_S f_n^{+} d\mu = \int_S f^{+} d\mu$$

$$\lim_{n \to \infty} \int_S f_n^{-} d\mu = \int_S f^{-} d\mu$$

What I tried:

Fatou's Lemmas for $f_n^{+}$

$$\int_S \limsup f_n^{+} d\mu \ge \limsup \int_S f_n^{+} d\mu \ge \liminf \int_S f_n^{+} d\mu \ge \int_S \liminf f_n^{+} d\mu$$

And that's about it. I have no idea if

$$\lim_{n \to \infty} f_n^{+}(s) = f^{+}(s) \forall s \in S$$

or a.e. in S.

Is

$$\lim_{n \to \infty} \max(f_n, 0) = \max(\lim_{n \to \infty} f_n, 0)$$

?

I seem to recall from basic calculus that

$$\lim_{x \to \infty} f(g(x)) = f(\lim_{x \to \infty} g(x))$$

if $f$ is continuous.

Even if it was true, I'm not sure what I can use here. I don't think I can use monotone convergence theorem or dominated convergence theorem. Can I?

How else can I approach this?

  • 1
    One direction of the equivalence is imediate. (=>) Suppose $\lim_{n \to \infty} \int_S |f_n - f| d\mu = 0$. Then since $$0 \leqslant \left |\int_S f_n - f d\mu \right| \leqslant \int_S |f_n - f| d\mu $$ we have $\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu$. – Ramiro Oct 27 '15 at 12:53
  • 1
    For the other direction, it is simpler to prove it using Lebesgue Dominated Convergence Theorem. (<=) Suppose $\lim_{n\to\infty}\int_S f_n d\mu = \int_S f d\mu$. Since $\{f_n\}$ is a sequence of nonnegative Lebesgue integrable functions and $\{f_n\}$ converges to function $f$ a.e., we have that $f\geqslant 0$ a.e.. So, without loss of generality, we can assume that $f\geqslant 0$. Now consider $f_n\wedge f$ defined by $(f_n\wedge f)(x)=\min\{f_n(x),f(x)\}$, for each $x\in\Omega$. Since $\{f_n\}$ converges to $f $ a.e., we have that $\{f_n\wedge f\}$ converges to $f$ a.e.. (to be continued) – Ramiro Oct 27 '15 at 12:56
  • 1
    Since, for all $n$, $\vert f_n\wedge f \vert = f_n\wedge f\leqslant f$ and $\int_{\Omega}f d\mu< \infty$, we can apply Lebesgue Dominated Convergence Theorem and we get $$\lim_{n\to\infty}\int_{\Omega}f_n\wedge f d\mu = \int_{\Omega}f d\mu$$ Now, note that $$\vert f_n-f\vert = f_n+f-2(f_n\wedge f)$$ So $$\int_{\Omega} \vert f_n-f\vert d\mu = \int_{\Omega}f_n d\mu +\int_{\Omega}f d\mu -2\int_{\Omega}(f_n\wedge f) d\mu$$ And so, since $\lim_{n \to \infty}\int_{\Omega } f_n d\mu = \int_{\Omega } f d\mu$, we have $$\lim_{n\to\infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$ Q.E.D. – Ramiro Oct 27 '15 at 13:10
  • 1
    The result you want to prove is a particular case of this question math.stackexchange.com/q/51502 (choose $p=1$). – saz Dec 1 '15 at 20:43
  • @saz I'm a dumbass. the proof was easy. thanks, and sorry for being an asshole. – BCLC May 3 at 15:13
up vote 5 down vote accepted
+50

The question has been edited. The updated question now asks to prove:

Suppose $\{f_n\}_{n \in \mathbb{N}}, f \in \mathscr{L}^1 (S, \Sigma, \mu)$ and $\lim_{n \to \infty} f_n(s) = f(s)$ a.e. in $S$. Then $\lim_{n \to \infty} \int_S |f_n - f| d\mu = 0$ iff $\lim_{n \to \infty} \int_S |f_n| d\mu = \int_S |f| d\mu$

Proof: (=>) It is trivial, since, from Minkowski's inequality, we have
$$\left | \int_S|f_n|d\mu-\int_S|f|d\mu \right |\leqslant \int_S|\,|f_n| -|f|\,| d\mu\leqslant \int_S|f_n -f| d\mu$$

(<=) Note that $|f_n -f|\leqslant |f_n| +|f|$. So, for each $n$, the function $|f_n| +|f| - |f_n -f|$ is non-negative and using Fatou's Lemma, we have \begin{align} 2 \int_S|f|d\mu &=\int_s \lim\inf(|f_n| +|f| - |f_n -f|)d\mu \leqslant \lim\inf \int_s (|f_n| +|f| - |f_n -f|)d\mu = \\ &=\lim\inf \left (\int_s|f_n|d\mu +\int_s|f|d\mu - \int_s|f_n -f|d\mu \right) = \\ &= \left(\lim\inf\int_s|f_n|d\mu\right) +\int_s|f|d\mu - \left(\lim\sup\int_s|f_n -f|d\mu\right) = \\ &=2\int_s|f|d\mu - \left(\lim\sup\int_s|f_n -f|d\mu\right) \end{align} So we have $$2 \int_S|f|d\mu \leqslant 2\int_s|f|d\mu - \left(\lim\sup\int_s|f_n -f|d\mu\right) $$ Since $f \in \mathscr{L}^1 (S, \Sigma, \mu)$ , we know that $\int_S|f|d\mu<+\infty$, and so we get $$\lim\sup\int_s|f_n -f|d\mu \leqslant 0$$ So we can conclude that $$\lim\int_s|f_n -f|d\mu = 0$$

Remark: There is another way to prove the (<=) part, which uses the Dominated Convergence Theorem (instead of Fatou's Lemma). However such way (for the question as currently stated) is a little bit "trickier" than the one presented above using Fatou's Lema. Here it is:

(<=) Consider $|f_n| \wedge |f|$ defined by $(| f_n | \wedge |f|)(x)=\min\{|f_n(x)|,|f(x)|\}$, for each $x \in \Omega$. Consider also $$ \sigma(fn,f)(x) = \left \{\begin{aligned} &= -1 &\textrm{ if } f_n(x)f(x)<0 \\ &= 0 &\textrm{ if } f_n(x)f(x)=0 \\ &= 1 &\textrm{ if } f_n(x)f(x)>0 \end{aligned}\right.$$ for each $x \in \Omega$.

Since $\{f_n\}$ converges to $f $ a.e., we have that $\{|f_n| \wedge |f|\}$ converges to $|f|$ a.e., and $\{\sigma(fn,f)\}$ converges to $\chi_{[f\neq 0]}$ a.e.. So, $\{\sigma(fn,f)(|f_n| \wedge |f|)\}$ converges to $|f|$ a.e.. But we know that, for all $n$, $\vert \sigma(fn,f)(|f_n| \wedge |f|) \vert = |f_n| \wedge |f| \leqslant |f| $ and $\int_{\Omega } |f| d\mu< \infty $. So we can apply Lebesgue Dominated Convergence Theorem and we have that $$\lim_{n \to \infty}\int_{\Omega } \sigma(fn,f)(|f_n| \wedge |f|) d\mu = \int_{\Omega } |f| d\mu$$ To conclute the proof, note that $$\vert f_n-f\vert = |f_n|+|f|-2\sigma(fn,f)(|f_n| \wedge |f|)$$ So $$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } |f_n| d\mu +\int_{\Omega } |f| d\mu -2\int_{\Omega } \sigma(fn,f)(|f_n| \wedge |f|) d\mu $$ And so, since $\lim_{n \to \infty}\int_{\Omega } |f_n| d\mu = \int_{\Omega } |f| d\mu$ and $\int_{\Omega } |f| d\mu<+\infty$, we have $$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$

  • Thanks Ramiro 1 Did you use Minkowski's for first ineq? 2 Did you mean function instead of functions? 3 Is the first = after 'we have' really supposed to be '='? – BCLC Dec 2 '15 at 14:34
  • Also what about the $f^{+}$ and $f^{-}$? – BCLC Dec 2 '15 at 14:39
  • 1
    @BCLC 1. Yes. In the (=>) part, (=>) $$\left | \int_S|f_n|d\mu-\int_S|f|d\mu \right |\leqslant \int_S|\,|f_n| -|f|\,| d\mu$$ is a corollary of Minkowski's inequality. 2. I meant functions (one for each $n$). Writing in clear way: "for each $n$, the function $|f_n| +|f| - |f_n -f|$ is non-negative". (I edited the answer to make this point clearer). 3. Yes, it is really "=". Please, note that $\lim\inf(|f_n| +|f| - |f_n - f|)= \lim(|f_n| +|f| - |f_n - f|)= 2|f|$ a.e.. – Ramiro Dec 2 '15 at 17:41
  • 1
    @BCLC No need to split $f_n$ and $f$ into their positive and negative parts. – Ramiro Dec 2 '15 at 17:44
  • 1
    @BCLC Please note that I added a remark to the answer, where I show another way to prove the (<=) part. – Ramiro Dec 3 '15 at 3:27

You're assuming that the $f_n$ are nonnegative, so $f_n^+ = f_n$ and you don't need to consider $f_n^-$.

First note that $$\left| \int_S f_n \, d\mu - \int_S f \, d\mu \right| \le \int_S |f_n - f| \, d\mu$$ so that one implication is easy and doesn't even require $f_n \to f$ a.e.

Next, since $|f_n - f| \le f_n + f$ you have $f + f_n - |f_n - f| \ge 0$ so that Fatou's Lemma is applicable. Since $f_n \to f$ a.e. you have $$ \int_S 2f \, d\mu = \int_S \liminf (f + f_n - |f_n - f|) \, d\mu \le \liminf \int_S f + f_n - |f_n - f| \, d\mu$$ where $$ \liminf \int_S f + f_n - |f_n - f| \, d\mu = \liminf \left[ \int_S f \, d\mu + \int_S f_n \, d\mu - \int_S |f_n - f| \, d\mu \right].$$ Next use a basic property of liminf: if $b_n \to b$, then $$\liminf (a + b_n - c_n) = a + b - \limsup c_n.$$ Thus \begin{align*}\int_S f_n \, d\mu \to \int_S f \,d\mu &\implies \liminf \int_S f + f_n - |f_n - f| \, d\mu = 2\int_S f \, d\mu - \limsup \int_S |f_n - f| \, d\mu \\ &\implies 2\int_S f \, d\mu \le 2\int_S f\, d\mu - \limsup \int_S |f_n - f| \, d\mu \\ &\implies \limsup \int_S |f_n - f| \, d\mu \le 0 \\ &\implies \int_S |f_n - f| \, d\mu \to 0.\end{align*}

  • Sorry Umberto P.! They're not nonnegative! Edited question. Already gave you an upvote – BCLC Nov 29 '15 at 23:54

It is simpler to prove Scheffe's lemma using Lebesgue Dominated Convergence Theorem.

(=>) Suppose $\lim_{n \to \infty} \int_S |f_n - f| d\mu = 0$. Then since $$0 \leqslant \left |\int_S f_n - f d\mu \right| \leqslant \int_S |f_n - f| d\mu $$ we have $\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu$.

(<=) Suppose $\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu$. Since $\{f_n\}$ is a sequence of nonnegative Lebesgue integrable functions and $\{f_n\}$ converges to function $f $ a.e., we have that $f\geqslant 0$ a.e.. So, without loss of generality, we can assume that $f\geqslant 0$.

Now consider $f_n \wedge f$ defined by $(f_n \wedge f)(x)=\min\{f_n(x),f(x)\}$, for each $x \in \Omega$.

Since $\{f_n\}$ converges to $f $ a.e., we have that $\{f_n \wedge f\}$ converges to $f$ a.e.. But we know that, for all $n$, $\vert f_n \wedge f \vert = f_n \wedge f \leqslant f $ and $\int_{\Omega } f d\mu< \infty $. So we can apply Lebesgue Dominated Convergence Theorem and we have that $$\lim_{n \to \infty}\int_{\Omega } f_n \wedge f d\mu = \int_{\Omega } f d\mu$$ To conclute the proof, note that $$\vert f_n-f\vert = f_n+f-2(f_n\wedge f)$$ So $$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } f_n d\mu +\int_{\Omega } f d\mu -2\int_{\Omega } (f_n\wedge f) d\mu $$ And so, since $\lim_{n \to \infty}\int_{\Omega } f_n d\mu = \int_{\Omega } f d\mu$ we have $$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$

  • Sorry Ramiro! They're not nonnegative! Edited question. Already gave you an upvote – BCLC Nov 29 '15 at 23:54

As stated in William's book, the functions $f_n$ are nonnegative, hence the only thing that you need to show is that $$\lim_{n \to \infty}\int f_n\,d\mu = \int f\,d\mu.$$ (Since $f$ is the pointwise limit of nonnegative functions, you can also conclude that $f$ is nonnegative almost everywhere.)

To prove this you only need the triangle inequality, indeed $$\lim_{n \to \infty}\Big|\int f_n\,d\mu - \int f\,d\mu\Big| \le \lim_{n \to \infty}\int |f_n - f|\,d\mu = 0.$$

As a side note, it is true that if $f_n \to f$ pointwise a.e. then $f_n^+ \to f^+$ pointwise a.e. and the reason is the one you stated: you can move the limit inside the maximum since $x \mapsto \max\{x,0\}$ is a (Lipschitz) continuous map.

  • Sorry Giovanni! They're not nonnegative! Edited question. Already gave you an upvote – BCLC Nov 29 '15 at 23:54

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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