If the limit of the integral exists, does the sequence of $L^1$-norms admit a convergent subsequence? Hi everyone: Let $ B $ be an open ball in $ \mathbb{R}^{n} $, $ n\geq2 $, and let $ (f_{n}) $ be a sequence of continuous (or even differentiable) functions such that 
$$ \int_{B}f_{n}(x)dx =-1$$ for all natural number $ n. $ Here is my question: can we conclude that there is a subsequence $ f_{n_{k}} $  such that
$$ \lim_{k\rightarrow+\infty}\int_{B}\lvert f_{n_{k}}(x)\rvert dx $$ exists?
 A: EDIT:  the proof of the counterexample below is in error, since the inference that is italicized in the last paragraph is false.  I suspect that this is still a valid counterexample, but I have not been able to correct the proof as of yet.

No.  For an explicit counter-example, first consider the series of functions $g_n$ on the unit ball in $\mathbb{R}^2$, specified in polar coordinates:
$$
g_n(r,\theta) = \frac{r \cos \theta}{r^3 + 1/n}.
$$
The integrand of $g_n(r,\theta)$ over the unit disc is always zero.  But the integrand of $|g_n(r,\theta)|$ over the unit disc is
$$
\iint_B |g_n(r,\theta)| = \int_0^{2\pi} \int_0^1 \frac{r^2 | \cos \theta | }{r^3 + 1/n} \, dr \,  d\theta = \frac{4}{3} \ln( n + 1 )
$$
which diverges monotonically as $n \to \infty$.  Note also that $g_n$ is strictly negative on the half-disc with $x < 0$, and the integral of $|g_n|$ on this half-disc is $\frac{2}{3} \ln (n + 1)$.
Now consider the functions 
$$
f_n(r, \theta) = g_n(r, \theta) - \frac{1}{\pi}.
$$
These obviously have the property that $\iint_B f_n \, dx = -1$.  But the integral on the half-disc with $x < 0$ satisfies
\begin{align*}
\iint_{B_{x<0}} |f_n(r,\theta)| \, dx &= \iint_{B_{x<0}} \left| g_n(r, \theta) - \frac{1}{\pi} \right| \, dx \\
&= \iint_{B_{x<0}} \left| g_n(r,\theta) \right| \, dx + \iint_{B_{x<0}} \left| -\frac{1}{\pi} \right|  \, dx \\
&= \frac{2}{3} \ln (n + 1) + 1,
\end{align*}
which is a monotone increasing unbounded sequence as $n \to \infty$.  The integral of $|f_n|$ over the other half of the disc is bounded below by zero;  thus, the integral of $|f_n|$ over the entire disc is also a monotone increasing unbounded sequence as $n \to \infty$ (being the sum of a monotone increasing unbounded sequence and a sequence that is bounded below.)  Finally, a monotone unbounded sequence cannot have a convergent subsequence;  thus, we have shown that the family of functions $f_n(r, \theta)$ given above is a counterexample.
A: $\newcommand{abs}[1]{\left\lvert{#1}\right\rvert}$Consider this lemma:

Let $B\subseteq \Bbb R^m$, $m\ge1$ be a non-empty open set and $f\in C^\infty_c(B)$. Then, there is $g\in C^\infty_c(B)$ such that $$\begin{cases}\int_{B} g(x)\,dx=\int_B f\,dx\\ \int_B \abs{g(x)}\,dx=1+\int_B\abs{f(x)}\,dx\end{cases}$$

Proof: Let $K=\operatorname{supp}f$. Since $K\subseteq B$ is compact, $B\setminus K$ is a non-empty open set and $C^\infty_c(B\setminus K)\subseteq C^\infty_c(B)$. Now, consider $h\in C^\infty_c(B\setminus K)\setminus\{0\}$ such that $$\int_B h(x)\,dx=0$$ Up to considering $\dfrac{1}{\int_B\abs{h(x)}\,dx}h$ instead of $h$, we can further require $$\int_B\abs{h(x)}\,dx=1$$  
Since $\operatorname{supp} g\cap\operatorname{supp}f=\emptyset$, we have $\abs{f+g}=\abs f+\abs g$.
So, $g=f+h$ is a function on $C^\infty_c(B)$ such as in the thesis of the lemma.

Applying consecutively the lemma to any function $f_0\in C^\infty_c(B)$ such that $\int_B\abs{f_0(x)}\,dx=-1$, you get a sequence of functions $f_n\in C^\infty_c(B)$ such that $$\int_B f_n(x)\,dx=-1\\\int_B\abs{f_n(x)}\,dx=n+\int_B\abs{f_0(x)}\,dx$$
And the last one is a strictly increasing divergent sequence: it does not have convergent subsequences.
A: This is not true. By translation and dilation, we might as well take the ball to be the ball of volume $1$ centered at the origin. Say that the functions $f_n$ are radially symmetric, so that they are only a function of distance from the origin. Then we can write $f_n(x)=g_n(r)$, where the $g_n$'s are functions of one variable. Choose them as
$$
g_n(r)=2nr-n-1.
$$
It follows that the $f_n$'s are infinitely differentiable functions whose integrals over the aforementioned ball are all $-1$. However,
$$
\int\limits_B|f_n|\,dx\geq\int\limits_{1/2}^1(2nr-n+1)\,dr\nearrow+\infty
$$
so that no subsequence can converge.
A: Counterexample: For the unit ball $B$ of $\mathbb R^k$ consider the functions $f_n((x_1,\dots, x_k)) = -1/V_k(B)+nx_1.$ Since $x_1$ is an odd function, $\int_Bx_1 = 0,$ hence $\int_B f_n =-1$ for all $n.$ But notice
$$\int_B |f_n| \ge \int_B (|nx_1| - |1/V_k(B)|) \to \infty.$$
A: Certainly not, and this has nothing to do with dimension $\ge 2$. One easily finds counterexamples for the one-dimensional case and can consider polar coordinates. (One may want to ensure that the functions vanish to sufficiently high order at $x=0$ in order for the polar coordinate trick to work)
