Doubt in Scheffe's Lemma

While reading up on "Glivenko Cantelli Theorem" from Probability Models by K.B Athreya, the author used 2 lemmae to prove it. One was called Scheffe's lemma, the other Polya's theorem.

Scheffe's Lemma is stated as follows:

Let $f_n, f$ be non negative $\mu$ integrable functions. If $f_n \to f$ a.e and $\int f_n d\mu \to \int fd\mu$, then $$\int |f_n - f|d\mu \to 0$$

My proof is:

Let $g_n = |f_n -f|$. Now we have $g_n \to 0$ a.e. Now $$0 \leq g_n = |f_n -f| \leq f + f_n$$ $$\Rightarrow \int g_n d\mu \leq \int fd\mu + \int f_nd\mu < \infty$$

Thus by Dominated convergence theorem, $$\int g_nd\mu \to 0$$ QED.

The Question: My doubt is that in this proof, I have not used that $\int f_n d\mu \to \int fd\mu$, at least not explicitly. So is this condition superfluous?

What I searched: I searched for Scheffe on MSE, but got 3 results (none useful) and when I typed Scheffe's instead, I got a result not belonging to these 3 which was actually on Scheffe's lemma (Though not helpful). It's strange (not the result but the search).

I'd appreciate any help/hints on this. Kindly ask me for clarifications if required.

• A function sequence cannot serve as a dominating function unless they also have a common dominating function. To circumvent this issue we need to use the assumption. Apr 17 '13 at 4:17
• Thanks. Could you please write it as an answer, I'd like to credit you. Apr 17 '13 at 4:20
• the last step of yours is to apply the dominated convergence theorem, in fact it is true, see the generalised version, for example, exercise 7.2. page 56 of Richard Bass' book (= Real analysis for graduate students, second edition, available on his homepage.). In your case, take $g_n = \vert f_n - f \vert$, $h_n = f_n + f$, then $\vert g_n \vert \leq h_n$, the other conditions can be easily verified. Oct 5 '14 at 10:39

Let

$$u_n = \max\{ f, f_n\} \quad \text{and} \quad l_n = \min \{ f, f_n \}$$

so that both $(u_n)$ and $(l_n)$ converge pointwise to $f$, $l_n \leq f \leq u_n$ and $|f - f_n| = u_n - l_n$. By DCT, it is clear that $\int l_n \, d\mu \to \int f \, d\mu$. thus from

$$\int u_n \, d\mu = \int (f + f_n - l_n) \, d\mu ,$$

taking $n\to\infty$, we have $\int u_n \, d\mu \to \int f \, d\mu$. (Here the assumption that you are concerning is used.) Therefore we have

$$\int |f - f_n| \, d\mu = \int u_n \, d\mu - \int l_n \, d\mu \to 0.$$

• very nice proof! Apr 17 '13 at 4:59
• Good use of $u_n$ and $l_n$. Apr 17 '13 at 5:02

Observe $|f| + |f_n| - |f-f_n|\geq 0$, with Fatou's lemma $$\liminf \int |f| + |f_n| - |f-f_n| \geq 2 \int |f|$$ by the assumption $\lim \int |f_n| = \int |f|$, the left-hand-side of above inequality also equals $$\liminf \int |f| + |f_n| - |f-f_n| = 2\int |f| -\limsup \int |f-f_n|,$$ combine them we get $$2\int |f| -\limsup \int |f-f_n|\geq 2 \int |f|$$ $$\limsup \int |f-f_n| \leq 0 .$$

• You are welcome. And similarly you can easily extend the result to $L^p$ using convexity that $$\bigg|\frac{f}{2}-\frac{f_n}{2}\bigg|^p \leq \frac{|f|^p}{2} + \frac{|f_n|^p}{2}$$ which implies $$0\leq \frac{|f|^p}{2} + \frac{|f_n|^p}{2}- \bigg|\frac{f}{2}-\frac{f_n}{2}\bigg|^p$$
– Xiao
Aug 30 '14 at 15:38
• @Xiao, Can I ask why $\displaystyle\liminf_n (|f|+|f_n|-|f-f_n|)=2|f|$ ? Thanks. Apr 10 '15 at 14:30
• @FardadPouran this is fatous lemma, $|f| + |f_n| - |f-f_n| \geq 0$ and $|f| + |f_n| - |f-f_n| \rightarrow 2|f|$ pointwise a.e.
– Xiao
Apr 11 '15 at 1:46
• Can we still say $\displaystyle\liminf_n (|f|+|f_n|+|f-f_n|)=2|f|$ in the case $f_n\xrightarrow{\mu}f$ rather than a.e ? Apr 11 '15 at 8:06
• Fatou's lemma will hold for $0 \leq f_n \rightarrow f$ in measure. But I dont think one can say $\liminf_n (|f|+|f_n| + |f-f_n|) = 2|f|$.
– Xiao
Apr 14 '15 at 0:13

I don't think your "proof" works. Let $f_{n}=n\chi[0,\frac{1}{n}]$, $f=0$. Then $f_{n}\rightarrow f$ almost everywhere. But you cannot conclude that $$\int f_{n}d\mu\rightarrow 0$$ as it is constant $1$. As the commenter pointed out to apply dominated convergence theorem you need a function $g$ such that $|f_{n}|\le |g|$ and $g\in L^{1}(\Omega)$.

• Indeed it is. Can't believe I made a rookie mistake. Anyhow well explained. Apr 17 '13 at 4:30