How to prove the 1-norm of a sequence of $L^1$ functions does not converge

Question

I've been learning Lebesgue Integral recently and the following is a problem in one of our previous sample exams:

If {$f_n$} is a bounded sequence of functions in $L^1(\mathbb{R})$, $f_n\to f$ a.e. and suppose that for any $M>0$, there exists $N\in\mathbb{N}$ such that $$\int_{\mathbb{R}\backslash[-M,M]}|f_N|>1$$ Prove that $||f_n||_1$ does NOT converge to $||f||_1$.

I tried to separate the integral $||f_n||_1$ into two parts: $[-M,M]$ and $\mathbb{R}\backslash[-M,M]$. The first part converges due to the Bounded Convergence Theorem. So it suffices to prove the second part does NOT converge. But I don't know where to go from there.

I also thought of functions, e.g. $\chi_{[-n-1,-n]\cup[n,n+1]}$ that satisfy the given conditions but had no idea how to use the properties of those functions. Any hint will be much appreciated!

Answer 1

• First of all, we can notice, as pointed out by PhoemueX, that if $f_n\to f$ almost everywhere and $\lVert f_n\rVert_1\to\lVert f\rVert_1$, then $\lVert f_n-f\rVert_1\to 0$. Consequently, it suffices to prove that we do not have $\lVert f_n-f\rVert_1\to 0$.
• To do so, it suffices to find a increasing sequence of integers $(n_k)_{k\geqslant 1}$ such that $\lVert f_{n_k}-f\rVert_1\geqslant 1/2$ for each $k$.
• Let us construct this sequence. Fatou's lemma combined with boundedness in $L^1$ of $(f_n)$ imply that $f$ is integrable. As a consequence, there exists $M_1$ such that $\int_{\mathbb R\setminus [-M_1,M_1]}\lvert f\rvert\leqslant 1/2$.
• Let $n_1$ be such that $\int_{\mathbb R\setminus [-M_1,M_1]}\lvert f_{n_1}\rvert\gt 1$. Then $\int_{\mathbb R\setminus [-M_1,M_1]}\lvert f_{n_1}-f\rvert\geqslant 1/2$.
• Suppose now that $n_1<\dots<n_k$ and $M_1<\dots<M_k$ are such that $\int_{\mathbb R\setminus [-M_i,M_i]}\lvert f_{n_i}-f\rvert\geqslant 1/2$ for each $i\in\{1,\dots,k\}$. Let us find $M_{k+1}>M_k$ and $n_{k+1}>n_k$ such that $\int_{\mathbb R\setminus [-M_{k+1},M_{k+1}]}\lvert f_{n_{k+1}}-f\rvert\geqslant 1/2$. Pick $M_{k+1}>M_k$ such that for each $\ell\in\{1,\dots,n_k\}$, $\int_{\mathbb R\setminus [-M_{k+1},M_{k+1}]}\lvert f_\ell\rvert\leqslant 1$. Applying the assumption with $M=M_{k+1}$, we can find $N$ such that $\int_{\mathbb R\setminus [-M_{k+1},M_{k+1}]}\lvert f_N\rvert\gt 1$ and by construction, this $N$ is necessarily bigger than $n_k$, hence $n_{k+1}=N$ does the job.
• We thus found sequences $(M_k)_{k\geqslant 1}$ and $(n_k)_{k\geqslant 1}$ for which the inequality $\int_{\mathbb R\setminus [-M_k,M_k]}\lvert f_{n_k}-f\rvert\geqslant 1/2$ holds for each $k$. Then $$ \int_{\mathbb R}\lvert f_{n_k}-f\rvert\geqslant \int_{\mathbb R\setminus [-M_k,M_k]}\lvert f_{n_k}-f\rvert\geqslant 1/2 $$ allows to conclude.

Answer 2

I assume we're considering the complete Lebesgue measure, i.e. subsets of null sets are measurable. We prove the following claim.

Claim: Let $f:\mathbb R\rightarrow\mathbb R$, and for $n\in\mathbb Z_{>0}$ let $f_n:\mathbb R\rightarrow\mathbb R$ be measurable, with the following properties. We have $\liminf_{n\rightarrow\infty}\|f_n\|_1< \infty$, we have $f_n(x)\rightarrow f(x)$ almost everywhere, and for each $M\in\mathbb R_{\ge 0}$ there exists $n\in\mathbb Z_{>0}$ such that $\int_{\mathbb R\setminus[-M,M]}|f_n(x)|\mathrm dx>1$. Then we have $\lim_{n\rightarrow\infty}\|f_n\|_1\neq\|f\|_1$, should the left hand side be well-defined.

First, let's get rid of some technical obstacles. For this purpose replace $f$ in the claim with $g:\mathbb R\rightarrow\mathbb R_{\ge 0}$ and $f_n$ by $g_n:\mathbb R\rightarrow\mathbb R_{\ge 0}$ to obtain the claim for non-negative functions. If the claim holds, then so does the claim for non-negative functions since this is a special case. If the claim does not hold, then there exist $f$, $f_n$ with all properties and such that $\|f\|_1=\lim_{n\rightarrow\infty}\|f_n\|_1$. Set $g(x)=|f(x)|$, $g_n(x)=|f_n(x)|$, notice that $g_n(x)\rightarrow g(x)$ a.e. and that the other properties as well as $\|g\|_1=\lim_{n\rightarrow\infty}\|g_n\|_1$ trivially hold, so the claim for non-negative functions also does not hold. This shows that the two claims are equivalent, and we may hence restrict to non-negative functions.

Next, notice that $f$ being an a.e. pointwise limit of measurable functions is itself measurable. Let $\mathcal E$ be a measurable set such that $f_n(x)\rightarrow f(x)$ for $x\in\mathcal E$ and such that $\mathbb R\setminus\mathcal E$ is a null set. Let $g_n(x)=f_n(x)$, $g(x)=f(x)$ for $x\in\mathcal E$, and $g_n(x)=g(x)=0$ otherwise. Then $g_n$, $g$ are measurable. Clearly, the claim and the claim with $f_n$, $f$ replaced by $g_n$, $g$ are equivalent, so we may also assume that $f_n$ converges pointwise to $f$ (not only a.e.).

Let $f_n$ have the desired properties. Now, we extract an increasing injection $\nu:\mathbb Z_{>0}\rightarrow\mathbb Z_{>0}$ such that $\int_{\mathbb R\setminus[-n,n]}g_n(x)\mathrm dx>1$, where $g_n=f_{\nu(n)}$, as follows. For given $M$ let $N(M)$ be such that $\int_{\mathbb R\setminus[-M,M]}f_{N(M)}(x)\mathrm dx>1$. Using $\|f_n\|_1<\infty$ and $f_n\ge 0$, let $M(n)$ be such that $\int_{\mathbb R\setminus[-M(n),M(n)]}f_n(x)\mathrm dx<1$. Let $\nu(1)=N(1)$. For $n\in\mathbb Z_{>0}$ let $M=\max\{M(n):n\in\mathbb Z\cap[1,\nu(n)]\}\cup\{n+1\}$ and $\nu(n+1)=N(M)$. Notice that $\nu(n+1)>\nu(n)$ because $\int_{\mathbb R\setminus[-M,M]}f_n(x)\mathrm dx<1$ for all $n\le\nu(n)$ (since $M\ge M(n)$) and that $\int_{\mathbb R\setminus[-(n+1),n+1]}f_{\nu(n+1)}(x)\mathrm dx>1$ since $M\ge n+1$. Hence, the sequence $\{g_n\}$ has the desired properties. Now, it is straightforward to check that the original claim and the following claim are equivalent.

Claim: For $n\in\mathbb Z_{>0}$ let $f_n:\mathbb R\rightarrow\mathbb R_{\ge 0}$ be measurable, with the following properties. The (measurable) pointwise limit $f:\mathbb R\rightarrow\mathbb R_{\ge 0}$, $x\mapsto\lim_{n\rightarrow\infty}f_n(x)$, exists, we have $\liminf_{n\rightarrow\infty}\|f_n\|_1<\infty$ for all $n$, and we have $\int_{\mathbb R\setminus[-n,n]}|f_n(x)|\mathrm dx>1$ for all $n$. Then we have $\lim_{n\rightarrow\infty}\|f_n\|_1\neq\|f\|_1$, if the left hand side is well-defined.

Let $f_n$ have the desired properties. First, notice that by Fatou's Lemma we have $\|f\|_1\le\liminf_{n\rightarrow\infty}\|f_n\|_1<\infty$ (considering the restriction to the Borel algebra of the completion of the Borel algebra does no harm here, since the domains can be any measure spaces). So, let $M$ be sufficiently large such that $\int_{\mathbb R\setminus[-M,M]}f(x)\mathrm dx\le 1$. Further, let $g=\chi_{[-M,M]}f$ and $g_n=\chi_{[-M,M]}f_n$, then $g_n$ converges pointwise to $g$ and Fatou's Lemma yields $\|g\|_1\le\liminf_{n\rightarrow\infty}\|g_n\|_1$, which further gives $\|f\|_1<\|g\|_1+1\le\liminf_{n\rightarrow\infty}\|g_n\|_1+1\le\liminf_{n\rightarrow\infty}\|f_n\|_1$. Thus, we have even established the following, stronger result.

Corollary: For $n\in\mathbb Z_{>0}$ let $f_n:\mathbb R\rightarrow\mathbb R_{\ge 0}$ be measurable, with the following properties. The (measurable) pointwise limit $f:\mathbb R\rightarrow\mathbb R_{\ge 0}$, $x\mapsto\lim_{n\rightarrow\infty}f_n(x)$, exists, we have $\liminf_{n\rightarrow\infty}\|f_n\|_1<\infty$ for all $n$, and we have $\int_{\mathbb R\setminus[-n,n]}|f_n(x)|\mathrm dx>1$ for all $n$. Then we have $\|f\|_1<\lim_{n\rightarrow\infty}\|f_n\|_1$.

As clarified in the comments, the assumption $\liminf_{n\rightarrow\infty}\|f_n\|_1<\infty$ is crucial. Otherwise, we may consider $f_n=\chi_{[-(n+1),n+1]}$ and $f\equiv 1$, yielding $\|f\|_1=\infty=\lim_{n\rightarrow\infty}\|f_n\|_1$. On the other hand, notice that we may even have $\|f_n\|_1=\infty$ for infintely many $n$, as long as pointwise convergence is preserved and we have $\|f_n\|_1<\infty$ for infintely many $n$. An example would be $f_{2n}=f+\chi_{\mathbb R\setminus[-2n,2n]}$, $f_{2n+1}=f+\chi_{[2n+1,2n+2]}+\chi_{[-2n-2,-2n-1]}$.

Answer 3

If the claim is false, then $f_n$ converges in $L^1.$ That is,

$\int_{\mathbb R}|f_n-f|\to 0.$

For each integer $M$, there is an integer $n(M)$ such that

$\int_{\mathbb{R}\backslash[-M,M]}|f_{n(M)}|>1.$

Then,

$\int_{\mathbb{R}\backslash[-M,M]}|f|+\int_{\mathbb{R}\backslash[-M,M]}|f_{n(M)}-f|>1$

and so in fact,

$\int_{\mathbb{R}\backslash[-M,M]}|f|+\int_{\mathbb R}|f_{n(M)}-f|>1.$

But $f$ is $L^1$ (why?) so if we can choose a subsequence of integers, $(M_k)_k$ such that $n(M_{k+1})>n(M_k)$ if we then let $k\to \infty,$ we will have a contradiction.

So, suppose we have chosen the first $k$ elements in such a way that $n(M_j)>n(M_{j-1}):\ 1\le j\le k.$ Since the $f_n$ are integrable, there is an integer $M'$ such that $\int_{\mathbb R\setminus [-M',M]}|f_m|<\frac{1}{2}$ for $1\le m\le n(M_k).$ But by hypothesis, for this $M'$, there is a $f_{n(M')}$ such that $\int_{\mathbb R\setminus [-M',M]}|f_{n(M')}|>1.$ We may take $M'>M_k.$ (why?). Then, $n(M')>n(M_k)$ so setting $M_{k+1}=M'$ completes the induction, and the proof.

Answer 4

As discussed in my other answer, we may restrict to sequences $f_n:\mathbb R\rightarrow\mathbb R_{\ge 0}$ of measurable functions such that the pointwise limit $f=\lim_{n\rightarrow\infty}f_n$ exists (with $\|f\|_\infty<\infty$), and such that $\int_{\mathbb R\setminus[-n,n]}f_n(x)\mathrm dx>1$. Now, we replace the assumption $\liminf_{n\rightarrow\infty}\|f_n\|_1<\infty$ by the following assumption.

Claim: Let $f_n$ be as described above, and further assume that $\liminf_{n\rightarrow\infty}\|f_n\|_\infty<\infty$. Then we have $\|f\|_1=\infty$ or $\|f\|_1<\liminf_{n\rightarrow\infty}\|f_n\|_1$.

For $\|f\|_1=\infty$ we are done, so let $\|f\|_1<\infty$. Then there exists $M$ such that $\int_{\mathbb R\setminus[-M,M]}f(x)\mathrm dx<1$. Also, there exists a subsequence $g_n$ of $f_n$ with $\|g_n\|<B$ for all $n$, where $B=2\liminf_{n\rightarrow\infty}\|f_n\|_\infty$ (obtained from the minimizing sequence for the limes inferior). Clearly, we still have $g_n\rightarrow f$ and $\int_{\mathbb R\setminus[-n,n]}g_n(x)\mathrm dx>1$ (since $g_n=f_N$ for some $N\ge n$). Now, the Bounded Convergence Theorem shows that $\|h_n\|_1\rightarrow\|h\|_1$ for $h_n=\chi_{[-M,M]}g_n$ and $h=\chi_{[-M,M]}f$. But this gives $\|f\|_1<\|h\|_1+1=\lim_{n\rightarrow\infty}\|h_n\|_1+1\le\lim_{n\rightarrow\infty}\|g_n\|_1$.

As discussed in the comments and the previous answer, the case $\|f\|_1=\infty$ is relevant since we may take $\chi_{[-n,n]}$. But if $\|f\|_1$ is finite, it is strictly lower than the limes inferior. As for the other answer, notice that we may even have $\|f_n\|_\infty=\infty$ for infinitely many $n$, and in particular we do not need uniform bounds (in $n$) for $\|f_n\|_\infty$. Also notice that the two results do not imply each other. The result in the other answer allows for the sequence $f_n=f+n\chi_{[-1/n,1/n]}+\chi_{[n,n+2]}$ which violates the assumptions here. Conversely, here we may consider $f_n=f+\chi_{[n,\infty)}$ (using that $\|f\|_\infty\le\liminf_{n\rightarrow\infty}\|f_n\|_\infty<\infty$) which violates the assumptions of the result in the other answer.