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Bounty Ended with 100 reputation awarded by Nate River
Corrected a stupid mistake
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fedja
fedja
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Let $$ J_\varepsilon=\inf_{h\in X_\varepsilon}\int\Phi(h)\,. $$ Clearly, $J_\varepsilon$ is a bounded non-increasing function on $(0,1)$, say. Let $J$ be its limit at $0+$. Passing to an appropriate subsequencedecreasing sequence $\varepsilon_n\to 0+$ and re-enumerating $X_n=X_{\varepsilon_n}, J_n=J_{\varepsilon_n}$, we can assume that $J_n\ge J-2^{-5n}$ We will choose a representative $h_n$ of $X_n$ for which $\int\Phi(h_n)$ is not more than $2^{-5n}$ above $J$ and, thereby, not more than $2\cdot 2^{-5n}$ above its infimum over $X_n$. Then for $m>n$, we have $h_{n,m}=\frac{h_n+h_m}2\in X_n$ (convexity of $X_n$ plus inclusion $X_n\supset X_m$) and $$ \int\Phi(h_{n,m})\le \int\frac 12(\Phi(h_n)+\Phi(h_m))-\int\frac{(h_n-h_m)^2}{4\pi (1+|h_n|^{2}+|h_m|^{2})} $$ (second order Taylor with the remainder in the Lagrange form), whence $$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2\cdot 2^{-5n} $$ regardless of $m>n$ (otherwise we would go below $J+2^{-5n}-2\cdot 2^{-5n}=J-2^{-5n}$, which is below the infimum over $X_n$). It remains to note that if $|h_n|\le 2^n$ and $|h_n-h_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get $$ m(\{|h_n|\le 2^n, |h_n-h_m|>2^{-n}\})\le \rm{Const}\,2^{-n} $$ for all $m>n$ from where it follows at once that $h_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h_n\|_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).

Let $$ J_\varepsilon=\inf_{h\in X_\varepsilon}\int\Phi(h)\,. $$ Clearly, $J_\varepsilon$ is a bounded non-increasing function on $(0,1)$, say. Let $J$ be its limit at $0+$. Passing to an appropriate subsequence $\varepsilon_n\to 0+$ and re-enumerating $X_n=X_{\varepsilon_n}, J_n=J_{\varepsilon_n}$, we can assume that $J_n\ge J-2^{-5n}$ We will choose a representative $h_n$ of $X_n$ for which $\int\Phi(h_n)$ is not more than $2^{-5n}$ above $J$ and, thereby, not more than $2\cdot 2^{-5n}$ above its infimum over $X_n$. Then for $m>n$, we have $h_{n,m}=\frac{h_n+h_m}2\in X_n$ (convexity of $X_n$ plus inclusion $X_n\supset X_m$) and $$ \int\Phi(h_{n,m})\le \int\frac 12(\Phi(h_n)+\Phi(h_m))-\int\frac{(h_n-h_m)^2}{4\pi (1+|h_n|^{2}+|h_m|^{2})} $$ (second order Taylor with the remainder in the Lagrange form), whence $$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2\cdot 2^{-5n} $$ regardless of $m>n$ (otherwise we would go below $J+2^{-5n}-2\cdot 2^{-5n}=J-2^{-5n}$, which is below the infimum over $X_n$). It remains to note that if $|h_n|\le 2^n$ and $|h_n-h_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get $$ m(\{|h_n|\le 2^n, |h_n-h_m|>2^{-n}\})\le \rm{Const}\,2^{-n} $$ for all $m>n$ from where it follows at once that $h_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h_n\|_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).

Let $$ J_\varepsilon=\inf_{h\in X_\varepsilon}\int\Phi(h)\,. $$ Clearly, $J_\varepsilon$ is a bounded non-increasing function on $(0,1)$, say. Let $J$ be its limit at $0+$. Passing to an appropriate decreasing sequence $\varepsilon_n\to 0+$ and re-enumerating $X_n=X_{\varepsilon_n}, J_n=J_{\varepsilon_n}$, we can assume that $J_n\ge J-2^{-5n}$ We will choose a representative $h_n$ of $X_n$ for which $\int\Phi(h_n)$ is not more than $2^{-5n}$ above $J$ and, thereby, not more than $2\cdot 2^{-5n}$ above its infimum over $X_n$. Then for $m>n$, we have $h_{n,m}=\frac{h_n+h_m}2\in X_n$ (convexity of $X_n$ plus inclusion $X_n\supset X_m$) and $$ \int\Phi(h_{n,m})\le \int\frac 12(\Phi(h_n)+\Phi(h_m))-\int\frac{(h_n-h_m)^2}{4\pi (1+|h_n|^{2}+|h_m|^{2})} $$ (second order Taylor with the remainder in the Lagrange form), whence $$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2\cdot 2^{-5n} $$ regardless of $m>n$ (otherwise we would go below $J+2^{-5n}-2\cdot 2^{-5n}=J-2^{-5n}$, which is below the infimum over $X_n$). It remains to note that if $|h_n|\le 2^n$ and $|h_n-h_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get $$ m(\{|h_n|\le 2^n, |h_n-h_m|>2^{-n}\})\le \rm{Const}\,2^{-n} $$ for all $m>n$ from where it follows at once that $h_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h_n\|_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).

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Corrected a stupid mistake
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fedja
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Let $I$ be the infimum of our functional. For every $n$$\varepsilon>0$ consider the set $X_n$$X_\varepsilon$ of all $L^1$-functions $h$ for which the value of the functional is at most $I+\frac 1n$$I+\varepsilon$. Clearly, it is a convex, non-empty, closed (in $L^1$) set and $X_1\supset X_2\supset\dots$. We need to choose some good representative in$X_{\varepsilon'}\supset X_{\varepsilon''}$ when $X_n$$\varepsilon'\ge\varepsilon''$. To this end, fix

Fix some strictly convex non-negative function $\Phi(t)\le |t|$. To simplify the technicalities, I'll choose it by the conditions $\Phi(0)=\Phi'(0)=1$, $\Phi''(t)=\frac 2{\pi(1+|t|^2)}$ but pretty much any other choice will work as well. We

Let $$ J_\varepsilon=\inf_{h\in X_\varepsilon}\int\Phi(h)\,. $$ Clearly, $J_\varepsilon$ is a bounded non-increasing function on $(0,1)$, say. Let $J$ be its limit at $0+$. Passing to an appropriate subsequence $\varepsilon_n\to 0+$ and re-enumerating $X_n=X_{\varepsilon_n}, J_n=J_{\varepsilon_n}$, we can assume that $J_n\ge J-2^{-5n}$ We will choose a representative $h_n$ of $X_n$ for which $\int\Phi(h_n)$ is not more than $2^{-5n}$ above $J$ and, thereby, not more than $2\cdot 2^{-5n}$ above its infimum over $X_n$. Then for $m>n$, we have $h_{n,m}=\frac{h_n+h_m}2\in X_n$ (convexity of $X_n$ plus inclusion $X_n\supset X_m$) and $$ \int\Phi(h_{n,m})\le \int\frac 12(\Phi(h_n)+\Phi(h_m))-\int\frac{(h_n-h_m)^2}{4\pi (1+|h_n|^{2}+|h_m|^{2})} $$ (second order Taylor with the remainder in the Lagrange form), whence $$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2^{-5n} $$$$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2\cdot 2^{-5n} $$ regardless of $m>n$ (otherwise we would go below $J+2^{-5n}-2\cdot 2^{-5n}=J-2^{-5n}$, which is below the infimum over $X_n$). It remains to note that if $|h_n|\le 2^n$ and $|h_n-h_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get $$ m(\{|h_n|\le 2^n, |h_n-h_m|>2^{-n}\})\le \rm{Const}\,2^{-n} $$ for all $m>n$ from where it follows at once that $h_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h_n\|_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).

Let $I$ be the infimum of our functional. For every $n$ consider the set $X_n$ of all $L^1$-functions $h$ for which the value of the functional is at most $I+\frac 1n$. Clearly, it is a convex, non-empty, closed (in $L^1$) set and $X_1\supset X_2\supset\dots$. We need to choose some good representative in $X_n$. To this end, fix some strictly convex non-negative function $\Phi(t)\le |t|$. To simplify the technicalities, I'll choose it by the conditions $\Phi(0)=\Phi'(0)=1$, $\Phi''(t)=\frac 2{\pi(1+|t|^2)}$ but pretty much any other choice will work as well. We will choose a representative $h_n$ of $X_n$ for which $\int\Phi(h_n)$ is not more than $2^{-5n}$ above its infimum over $X_n$. Then for $m>n$, we have $h_{n,m}=\frac{h_n+h_m}2\in X_n$ (convexity of $X_n$ plus inclusion $X_n\supset X_m$) and $$ \int\Phi(h_{n,m})\le \int\frac 12(\Phi(h_n)+\Phi(h_m))-\int\frac{(h_n-h_m)^2}{4\pi (1+|h_n|^{2}+|h_m|^{2})} $$ (second order Taylor with the remainder in the Lagrange form), whence $$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2^{-5n} $$ regardless of $m>n$. It remains to note that if $|h_n|\le 2^n$ and $|h_n-h_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get $$ m(\{|h_n|\le 2^n, |h_n-h_m|>2^{-n}\})\le \rm{Const}\,2^{-n} $$ for all $m>n$ from where it follows at once that $h_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h_n\|_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).

Let $I$ be the infimum of our functional. For every $\varepsilon>0$ consider the set $X_\varepsilon$ of all $L^1$-functions $h$ for which the value of the functional is at most $I+\varepsilon$. Clearly, it is a convex, non-empty, closed (in $L^1$) set and $X_{\varepsilon'}\supset X_{\varepsilon''}$ when $\varepsilon'\ge\varepsilon''$.

Fix some strictly convex non-negative function $\Phi(t)\le |t|$. To simplify the technicalities, I'll choose it by the conditions $\Phi(0)=\Phi'(0)=1$, $\Phi''(t)=\frac 2{\pi(1+|t|^2)}$ but pretty much any other choice will work as well.

Let $$ J_\varepsilon=\inf_{h\in X_\varepsilon}\int\Phi(h)\,. $$ Clearly, $J_\varepsilon$ is a bounded non-increasing function on $(0,1)$, say. Let $J$ be its limit at $0+$. Passing to an appropriate subsequence $\varepsilon_n\to 0+$ and re-enumerating $X_n=X_{\varepsilon_n}, J_n=J_{\varepsilon_n}$, we can assume that $J_n\ge J-2^{-5n}$ We will choose a representative $h_n$ of $X_n$ for which $\int\Phi(h_n)$ is not more than $2^{-5n}$ above $J$ and, thereby, not more than $2\cdot 2^{-5n}$ above its infimum over $X_n$. Then for $m>n$, we have $h_{n,m}=\frac{h_n+h_m}2\in X_n$ (convexity of $X_n$ plus inclusion $X_n\supset X_m$) and $$ \int\Phi(h_{n,m})\le \int\frac 12(\Phi(h_n)+\Phi(h_m))-\int\frac{(h_n-h_m)^2}{4\pi (1+|h_n|^{2}+|h_m|^{2})} $$ (second order Taylor with the remainder in the Lagrange form), whence $$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2\cdot 2^{-5n} $$ regardless of $m>n$ (otherwise we would go below $J+2^{-5n}-2\cdot 2^{-5n}=J-2^{-5n}$, which is below the infimum over $X_n$). It remains to note that if $|h_n|\le 2^n$ and $|h_n-h_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get $$ m(\{|h_n|\le 2^n, |h_n-h_m|>2^{-n}\})\le \rm{Const}\,2^{-n} $$ for all $m>n$ from where it follows at once that $h_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h_n\|_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).

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fedja
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As it has been already noted in the comments, the minimizer doesn't need to be unique. However, it always exists. It is not terribly hard to show but it is not trivial either, so I wonder why the question attracted so few votes.

The proof consists of two independent parts. The first one is that the limit of every minimizing sequence that converges almost everywhere (or just in measure) is a minimizer and the second one is that there exists a minimizing sequence converging almost everywhere. I will use the fact that we deal with a finite measure space though it should, probably, be irrelevant. WLOG, we may assume that $\|f_j\|_1\le 1$ for all $j$.

Part 1

Assume that $h_n$ is a minimizing sequence and $h$ is its pointwise limit (or just limit in measure). Since we can assume WLOG that $\|h_n\|_1\le 3$ (to be a competitor, you need to perform not much worse than $0$, at the very least), we have $\|h\|_1\le 3$ as well (by Fatou). Then, by the definition of the convergence in measure, we can write $h_n=u_n+v_n$ where $u_n$ converge to $h$ uniformly and $v_n$ are supported on $E_n$ with $m(E_n)\to 0$ as $n\to\infty$. Also $\|v_n\|_1\le 7$, say.

Now, since $v_n\in L^1$, we can find $\delta_n>0$ such that for every set $E$ with $m(E)<\delta_n$, we have $\int_E|v_n|<\frac 1n$, say. By induction, we can now choose a subsequence $n_k$ such that $$ \sum_{q=k+1}^\infty m(E_{n_q})<\delta_{n_k}\,. $$ Then $\|v_{n_k}\chi_{\cup_{q>k}E_{n_q}}\|_1\to 0$ and, adding these parts to $u_{n_k}$, we see that we can (passing to a subsequence) assume that our minimizing sequence can be represented as $h_n=U_n+V_n$ where $U_n\to h$ in $L^1$ and $V_n$ have disjoint supports $G_n$.

Since correcting a minimizing sequence by a sequence tending to $0$ in $L^1$ results in a minimizing sequence, we can just as well assume that $h_n=h+V_n$.

If $\|V_n\|_1\to 0$ (even along a subsequence), we are done. Assume now that $\|V_n\|_1\ge \tau>0$. Then for every function $g$ of $L^1$-norm less than $4$, we have $$ \|g\|_1\le \max(\|g-V_n\|_1,\dots,\|g-V_{n+N-1}\|_1) $$ for all $n$ as soon as $N>8/\tau$. Indeed, since the supports $G_n$ of $V_n$ are disjoint, there is $q\in\{0,\dots,N-1)$ such that $\int_{G_{n+q}}|g|\le \frac 1N\|g\|_1< \frac\tau 2$, in which case subtracting $V_{n+q}$ can only drive the norm up.

Applying this to the functions $g_j=f_j-h$, we conclude that $$ \limsup_{j\to\infty}\|f_j-h\|_1\le \limsup_{j\to\infty}\max_{0\le q\le N-1}\|f_j-h_{n+q}\|_1= \max_{0\le q\le N-1}\limsup_{j\to\infty}\|f_j-h_{n+q}\|_1 $$ for every $n$, i.e. $h$ is a minimizer in this case as well.

Part 2

Let $I$ be the infimum of our functional. For every $n$ consider the set $X_n$ of all $L^1$-functions $h$ for which the value of the functional is at most $I+\frac 1n$. Clearly, it is a convex, non-empty, closed (in $L^1$) set and $X_1\supset X_2\supset\dots$. We need to choose some good representative in $X_n$. To this end, fix some strictly convex non-negative function $\Phi(t)\le |t|$. To simplify the technicalities, I'll choose it by the conditions $\Phi(0)=\Phi'(0)=1$, $\Phi''(t)=\frac 2{\pi(1+|t|^2)}$ but pretty much any other choice will work as well. We will choose a representative $h_n$ of $X_n$ for which $\int\Phi(h_n)$ is not more than $2^{-5n}$ above its infimum over $X_n$. Then for $m>n$, we have $h_{n,m}=\frac{h_n+h_m}2\in X_n$ (convexity of $X_n$ plus inclusion $X_n\supset X_m$) and $$ \int\Phi(h_{n,m})\le \int\frac 12(\Phi(h_n)+\Phi(h_m))-\int\frac{(h_n-h_m)^2}{4\pi (1+|h_n|^{2}+|h_m|^{2})} $$ (second order Taylor with the remainder in the Lagrange form), whence $$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2^{-5n} $$ regardless of $m>n$. It remains to note that if $|h_n|\le 2^n$ and $|h_n-h_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get $$ m(\{|h_n|\le 2^n, |h_n-h_m|>2^{-n}\})\le \rm{Const}\,2^{-n} $$ for all $m>n$ from where it follows at once that $h_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h_n\|_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).

That's it. Feel free to ask questions if anything is unclear :-)