Dominated Convergence Theorem

Give an example of a sequence $\{f_n\}_{n=1}^\infty$ of integrable functions on $\mathbb{R}$ such that $f_n \to f$ but $\int f_n \not\to \int f$. Explain why your example does not conflict with the Dominated Convergence Theorem.

I do notice that the inequality $|f_n(x)| \le g(x)$, where $g$ is an integrable function over $\mathbb{R}$, is not listed here in this problem. So the function need not be dominated by an integrable function here. But this is required as one hypothesis of the Dominated Convergence Theorem; hence the example will not conflict.

If this is sound reasoning, how may I come up with functions that are not dominated by another function? Initially I was thinking $f_n(x)=x \sin (nx)$ because its lim sup is $\infty$, but even then, we still have $|f_n(x)| \le |x| =: g(x)$.

• $f_n = n 1_{(0,{1 \over n}]}$. – copper.hat May 8 '15 at 16:35
• It is not only a matter of being dominated by a function, but by an integrable function. $|x|$ is not integrable on $\mathbb{R}$, as $\int_{-\infty}^\infty |x| dx = \infty$. – Pedro M. May 8 '15 at 16:39
• @PedroM. I forgot about the "integrable" function part. :O But thanks for clearing that up. – Cookie May 8 '15 at 16:42
• It may be dominated by another function, however the detail is that the if dominating function does not belong to $L_1$ then DCT may not hold. – Alonso Delfín May 8 '15 at 16:42
• @AaronMaroja I do not understand why you centerized "$\int f \not\to \int f$" and removed the period at the end of that sentence. – Cookie May 9 '15 at 7:28

Consider the sequence of functions on $(0,1)$ $$f_n(x) = \begin{cases} n & \text{ if } x \in (0,1/n)\\ 0 & \text{ otherwise} \end{cases}$$ We have $\lim_{n \to \infty} f_n(x) = 0 = f(x)$. However, $$\lim_{n \to \infty} \int_0^1f_n(x)dx = 1 \neq 0 = \int_0^1 f(x) dx$$ The key in the dominated convergence theorem is that sequence of functions $f_n(x)$ must be dominated by a function $g(x)$, which is also integrable.

I would like to provide a slightly more abstract framework to illustrate this "loss of compactness"$^{}$ phenomenon. The functional setting is the $L^1(\mathbb{R})$ space: \begin{equation} L^1(\mathbb{R})=\left\{f\colon\mathbb{R}\to\mathbb{R}\ :\ \int_{-\infty}^\infty \lvert f(x)\rvert\, dx<\infty \right\},\qquad \lVert f\rVert=\lVert f\rVert_{L^1}=\int_{-\infty}^\infty\lvert f(x)\rvert\, dx. \end{equation} Here we take a bounded sequence $f_n$ that converges pointwise a.e. : \begin{equation} \begin{array}{cc} \|f_n\|\le C, & f_n\to f\, \text{a.e.} \end{array} \end{equation} The question is: does this sequence converge, that is, is it true that \begin{equation} \lVert f_n-f\rVert\to 0?^{\ } \end{equation} This would make for a very desirable property of our functional space. However, as other answers very clearly show, the answer is negative in general. The problem is that our space is subject to the action of noncompact groups of isometries. Namely, one has the action of the translation group \begin{equation} \begin{array}{cc} \left(T_\lambda f\right)(x)=f(x-\lambda), &\lambda \in (\mathbb{R}, +) \end{array} \end{equation} and of the dilation group \begin{equation} \begin{array}{cc} \left(D_\lambda f\right)(x)=\frac{1}{\lambda}f\left(\frac{x}{\lambda}\right), &\lambda \in (\mathbb{R_{>0}}, \cdot) \end{array} \end{equation} The change of variable formula for integrals immediately shows that those group actions are isometric, that is, they preserve the norm.

So, fixing a non-vanishing function $f\in L^1(\mathbb{R})$, its orbits $T_\lambda f$ and $D_\lambda f$ form bounded and non-compact subsets of $L^1(\mathbb{R})$. In particular, letting $\lambda \to +\infty$ (or $\lambda \to -\infty$ for translations, or $\lambda\to 0$ for dilations), one finds counterexamples to the question above. (Note that, more or less, all the examples constructed in the other, excellent, answers are constructed this way).

In technical jargon one says that the translation and dilation groups introduce a defect of compactness in $L^1(\mathbb{R})$ space. This is the terminology of the Concentration-Compactness theory (the linked page is a blog entry of T. Tao, but the theory has been founded by P.L. Lions). The dominated convergence theorem can be therefore seen as a device that impedes the defect of compactness to take place.

Footnotes

$^{}$ The OP only asks about the convergence of the integrals: $\int f_n\to \int f$. Now a standard theorem (cfr. Lieb & Loss Analysis, 2nd ed. Theorem 1.9 (Missing term in Fatou's lemma), see in the remarks) gives us the equivalence \begin{equation} \begin{array}{ccc} \lVert f_n\rVert_{L^1} \to \lVert f\rVert_{L^1} & \iff & \lVert f_n-f\rVert_{L^1}, \end{array} \end{equation} Therefore, at least for sequence of positive functions for which $\int f_n=\lVert f_n\rVert_{L^1}$, the failure of convergence for sequences of integrals is exactly the same thing as the failure of convergence in $L^1$ space. That's why one can see the phenomenon in this functional analytic setting.

$^{}$ Compactness usually means that bounded sequences have convergent subsequences. In $L^1(\mathbb{R})$ space, pointwise convergent sequences are compact if and only if they are norm convergent.

• wow! Very interesting point of view, never heard of that before! :-) – Ant May 8 '15 at 18:47

It is not only a matter of being dominated by just any function $g$, but by an integrable function. $|x|$ is not integrable on $\mathbb{R}$, as $\int_{-\infty}^\infty |x| dx = \infty$.

copper.hat provided an example that is not dominated by any function, but you can also pick bounded examples, such as $f_n = 1_{[n,n+1]}$.

• Does $f_n=\chi_{[-n,n]}$ also work? (Sorry, I'm used to the characteristic function that uses the Greek letter $\chi$ instead of $1$.) – Cookie May 8 '15 at 16:46
• @dragon: Then $f_n$ converges to the constant function $1$, which is not integrable on $\mathbb{R}$. This is another illustration of how the DCT may fail if the dominating function $g$ is not integrable: $\int f = \infty$ (in my example, $\int f$ is finite but is different from $\lim \int f_n$). Notice, however, that in your case, $\int f_n \to \infty = \int f$. – Pedro M. May 8 '15 at 16:49

Let $$f_n(x) = \frac{n^3x^2}{1+n^4x^4}.$$ Then $f_n(x) \to 0$ pointwise everywhere, and $\int_\infty^\infty f_n(x)\,dx = \int_\infty^\infty \frac{x^2}{1+x^4}\,dx$ for every $n.$

Let $f_n = \chi[n,2n]$ then for any $x \in \mathbb R$

$$\lim_{n \to \infty} f_n (x) = 0 \,\,\, \text{and} \,\,\, \int_{x\in \mathbb R} f_n(x) dx = n \to \infty \,\,\, \text{as}\,\, n \to \infty$$

On the other hand taking $f(x) = 0 , \forall x \in \mathbb R$ we have $$\int_{x\in \mathbb R} f(x) dx = \int _{x\in \mathbb R} 0\,\, dx = 0$$ then

$$\int_{x\in \mathbb R} f_n(x) dx \not \to \int_{x\in \mathbb R} f(x) dx$$