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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?

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    $\begingroup$ 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 $\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu$. $\endgroup$
    – Ramiro
    Commented Oct 27, 2015 at 12:53
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    $\begingroup$ 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) $\endgroup$
    – Ramiro
    Commented Oct 27, 2015 at 12:56
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    $\begingroup$ 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. $\endgroup$
    – Ramiro
    Commented Oct 27, 2015 at 13:10
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    $\begingroup$ The result you want to prove is a particular case of this question math.stackexchange.com/q/51502 (choose $p=1$). $\endgroup$
    – saz
    Commented Dec 1, 2015 at 20:43
  • $\begingroup$ @saz I'm a dumbass. the proof was easy. thanks, and sorry for being an asshole. $\endgroup$
    – BCLC
    Commented May 3, 2018 at 15:13

5 Answers 5

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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 $$ \sigma(f_n,f)(x) = \left \{\begin{aligned} &= 0 &\textrm{ if } f_n(x)f(x)\leqslant 0 \\ %&= 0 &\textrm{ if } f_n(x)f(x)=0 \\ &= 1 &\textrm{ if } f_n(x)f(x)>0 \end{aligned}\right.$$ 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 $\{\sigma(f_n,f)\}$ converges to $\chi_{[f\neq 0]}$ a.e.. So, $\{\sigma(f_n,f)(|f_n| \wedge |f|)\}$ converges to $|f|$ a.e.. But we know that, for all $n$, $\vert \sigma(f_n,f)(|f_n| \wedge |f|) \vert \leqslant |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 $$\lim_{n \to \infty}\int_{\Omega } \sigma(f_n,f)(|f_n| \wedge |f|) d\mu = \int_{\Omega } |f| d\mu$$ To conclude the proof, note that $$\vert f_n-f\vert = |f_n|+|f|-2\sigma(f_n,f)(|f_n| \wedge |f|)$$ So $$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } |f_n| d\mu +\int_{\Omega } |f| d\mu -2\int_{\Omega } \sigma(f_n,f)(|f_n| \wedge |f|) d\mu $$ 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 $$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$

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  • $\begingroup$ 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 '='? $\endgroup$
    – BCLC
    Commented Dec 2, 2015 at 14:34
  • $\begingroup$ Also what about the $f^{+}$ and $f^{-}$? $\endgroup$
    – BCLC
    Commented Dec 2, 2015 at 14:39
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    $\begingroup$ @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.. $\endgroup$
    – Ramiro
    Commented Dec 2, 2015 at 17:41
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    $\begingroup$ @BCLC No need to split $f_n$ and $f$ into their positive and negative parts. $\endgroup$
    – Ramiro
    Commented Dec 2, 2015 at 17:44
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    $\begingroup$ @BCLC Please note that I added a remark to the answer, where I show another way to prove the (<=) part. $\endgroup$
    – Ramiro
    Commented Dec 3, 2015 at 3:27
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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*}

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  • $\begingroup$ Sorry Umberto P.! They're not nonnegative! Edited question. Already gave you an upvote $\endgroup$
    – BCLC
    Commented Nov 29, 2015 at 23:54
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Considering that $f$ is non-negative and, for each $n$, $f_n$ is non-negative, then 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 $$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$

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  • $\begingroup$ Sorry Ramiro! They're not nonnegative! Edited question. Already gave you an upvote $\endgroup$
    – BCLC
    Commented Nov 29, 2015 at 23:54
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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.

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  • $\begingroup$ Sorry Giovanni! They're not nonnegative! Edited question. Already gave you an upvote $\endgroup$
    – BCLC
    Commented Nov 29, 2015 at 23:54
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This is the same question I encountered while I was reading the textbook. For the nontrivial direction, I will try to give a proof following the textbook's approach, which is to prove $\mu(f_n^+) \rightarrow \mu(f^+)$ and $\mu(f_n^-) \rightarrow \mu(f^-)$ and then apply the same proof in the first part of Scheffe's lemma. But I don't whether it is rigorous or not.

First since $f_n \rightarrow f$ (a.e.), then $f_n^+ \rightarrow f^+$ and $f_n^- \rightarrow f^-$. Then since $f_n^+$ and $f_n^-$ are nonnegative functions, we can apply Fatou's Lemma respectively, $$ \mu(f^+) = \mu(liminf f_n^+) \leq liminf \mu(f_n^+)$$ $$ \mu(f^-) = \mu(liminf f_n^-) \leq liminf \mu(f_n^-)$$

Moreover, $$\text{limsup } \mu(f_n^+) = \text{limsup } \mu(|f|-f_n^-) \leq \text{limsup } \mu(|f|) - \text{liminf } (f_n^-) \leq \mu(|f|) -\mu(\text{liminf } f_n^-) = \mu(|f|) - \mu(f^-) = \mu(f^+) $$

These imply $$\text{limsup } \mu(f_n^+) \leq \mu(f^+) \leq \text{liminf } \mu(f_n^+)$$ which implies $$\mu(f_n^+) \rightarrow \mu(f^+)$$

Apply the same procedure to prove $\mu(f_n^-) \rightarrow \mu(f^-)$ and complete the proof by applying the same proof in the first part of Scheffe's lemma as stated in William's book.

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