Approximating $f \in L^1(M)$ with integral zero with $f_n \in L^\infty(M)$ with integral zero? Let $M$ be a closed Riemannian manifold and let $f \in L^1(M)$ with $\int_M f =0$. Is it possible to find a sequence $f_n \in L^\infty(M)$ with $\int_M f_n = 0$ such that $f_n \to f$ in $L^1(M)$?
I got a hint to use continuity of $t \mapsto \int_M \text{min}(g(x), t)$ and the intermediate value theorem but I didn't solve that.
 A: One idea to prove this proposition is the following :


*

*Choose $s > 0$. Cut off $f$ under $-s$. Get a function $g_s$.

*Use the hint. Find a $t > 0$ such that the cut-off of $g_s$ above $t$ has integral $0$. Get a function $f_s \in \mathbb{L}^\infty$ with integral $0$.

*Prove that $f_s$ converges to $f$ in $\mathbb{L}^1$ when $s$ goes to $+ \infty$.
Let's write things down. If $f$ is essentially bounded, there is nothing to prove (take $f_n = f$). If $f$ isn't essentially bounded, up to working with $-f$, we can assume without loss of generality that $Leb (f < -s) > 0$ for all $s$.
Fix $s > 0$. Let $g_s := \max \{-s, f\}$. Then $\|f-g_s\|_{\mathbb{L}^1} = \int_{\{f < -s\}} |f+s| \leq \|f\|_{\mathbb{L}^1}$, so $g_s \in \mathbb{L}^1$. In addition, $g_s-f \geq 0$ and is not identically $0$, so $\int_M g_s > 0$.
For all $T \geq 0$, let :
$$G(T) := \int_M \min \{g_s, T\} (x) dx.$$
Note that $|\min \{g_s, T\}| \leq |f|$ for all $T$, and that $\min \{g_s, S\}$ converges pointwise to $\min \{g_s, T\}$ when $S$ converges to $T$. By the dominated convergence theorem, $G$ is continuous. In addition, $G(0) \leq -s Leb(f < -s) < 0$ and $\lim_{+ \infty} G = \int_M g_s > 0$.
By the intermediate value theorem, there exists a $T$ such that $G(T) = 0$. Let us take, for instance, $T(s)$ as the minimum of such $T$'s. Let $f_s := \min \{g_s, T(s)\}$. Then :


*

*$-s \leq f_s \leq T(s)$, so $f_s \in \mathbb{L}^\infty$ for all $s$;

*$\int_M f_s = G(T(s)) = 0$;

*$\int_M |f-f_s| = 2 \int_M |f-g_s|$.
But $|g_s| \leq |f|$ and $(g_s)$ converges to $f$ pointwise. Again, by the dominated convergence theorem, $\lim_{s \to + \infty} \int_M |f-g_s| = 0$. Hence, $(f_s)$ converges to $f$ in $\mathbb{L}^1$ as $s$ goes to infinity.

The proof above is in the spirit of the hint you posted. However, I think we can get more efficient.
First, bounded measurable functions are dense in $\mathbb{L}^1$. Let $h$ be a bounded non-negative function with integral $1$ (take for instance $h(x) = 1_A (x) / Leb(A)$, where $A$ is any measurable set with positive and finite measure).
Let $\varepsilon > 0$. Take $g$ bounded such that $\|f-g\|_{\mathbb{L}^1} \leq \varepsilon$. Let $g_\varepsilon := g + (\int_M (f-g))h$. Then $g_\varepsilon$ is bounded, has integral $0$, and $\|f-g_\varepsilon \|_{\mathbb{L}^1} \leq 2\varepsilon$, which is what we want.
The advantage of this approch - besides its simplicity - is that you can do better than "bounded". For instance, you can use the fact that $\mathcal{C}_c^\infty (M, \mathbb{R})$ is dense in $\mathbb{L}^1$, and replace $h$ by a bump function. Then you can approximate $f$ by smooth functions with compact support, instead of merely bounded functions.
