Proof. We need only prove the case $n=2$ . Once we have proved the case $n=2$ , the general case follows by induction as follows. Suppose that the conclusion holds for $n=m-1$ . For $n=m$ , by the cases $n=2$ and $n=m-1$ ,

$$ \begin{align*} f\bigg(\sum_{k=1}^m \lambda_k x_k\bigg) & =f\bigg(\lambda_1x_1+(1-|\lambda_1|)\sum_{k=2}^m\frac{\lambda_k}{1-|\lambda_1|}x_k\bigg)\\ &= \lambda_1'f(x_1)+\lambda f\bigg(\sum_{k=2}^m\frac{\lambda_k}{1-|\lambda_1|}x_k\bigg)=\lambda_1'f(x_1)+\lambda \sum_{k=2}^m\frac{\lambda_k'}{1-|\lambda_1|} f(x_k), \end{align*} $$

where $|\lambda |=1-|\lambda _1|$ and $|\lambda _k'|=|\lambda _k|$ for $k=1,2,\ldots , n$ . This completes the induction proof.

For $n=2$ , since f is odd by Lemma 2.1, it is easy to check the cases $|\lambda _1|=0$ and $|\lambda _1|=1$ . Assume that $0<|\lambda _1|<1$ . Let $A_1=\operatorname {supp}(x_1)$ and $A_2=\operatorname {supp}(x_2)$ . By the hypothesis, $\mu (A_1\cap A_2)=0$ . For any choice of signs $\theta _1, \theta _2 \in \{-1,1\}$ , the min-phase-isometry property of f gives

$$ \begin{align*} &\min\{ \|\theta_1f(x_1)+\theta_2f(x_2)\|, \|\theta_1f(x_1)-\theta_2f(x_2)\|\}\\ &\quad = \min \{\|f(x_1)+ f(x_2)\|, \|f(x_1)-f(x_2)\|\}=\min \{\|x_1+x_2\|, \|x_1-x_2\|\}=2, \end{align*} $$

which implies

(2.1) $$ \begin{align} \|\theta_1f(x_1)+ \theta_2f(x_2)\|=\|\theta_1f(x_1)- \theta_2f(x_2)\|=2. \end{align} $$

Lemma 2.2 guarantees that

$$ \begin{align*}\|\theta_1\lambda_1 f(x_1)+\theta_2\lambda_2 f(x_2)\|=|\theta_1\lambda_1|+|\theta_2\lambda_2|=|\lambda_1|+|\lambda_2|=1.\end{align*} $$

By the surjectivity of f, there exists $z\in S_{L^1(\mu )}$ such that

(2.2) $$ \begin{align} f(z)=\theta_1\lambda_1 f(x_1)+\theta_2\lambda_2 f(x_2). \end{align} $$

Once we establish

(2.3) $$ \begin{align} z=\mu_1 x_1+ \mu_2 x_2 \quad\text{with } |\mu_k|=|\lambda_k|,\ k=1,2, \end{align} $$

Lemma 2.1 will imply that

$$ \begin{align*} \{\theta_1 \lambda_1 f(x_1)+ \theta_2\lambda_2 f(x_2): |\theta_1|=|\theta_2|=1\}=\{f(\theta_1 \lambda_1 x_1+ \theta_1\lambda_2 x_2): |\theta_1|=|\theta_2|=1\}, \end{align*} $$

which will complete the proof. Thus, we need only prove (2.3).

Suppose that

(2.4) $$ \begin{align} \min\{\|f(z)+ f(x_k)\|, \|f(z)- f(x_k)\|\}=\|z-\alpha_k x_k\| \end{align} $$

with $|\alpha _k|=1$ for $k=1,2$ . Define the sets

$$ \begin{align*} A_k^+:=\{t\in A_k: \alpha_k z(t)\cdot x_k(t)\geq 0\}\quad \text{and}\quad A_k^-:=\{t\in A_k: \alpha_k z(t)\cdot x_k(t)< 0\} \end{align*} $$

for $k=1,2$ . We claim that

(2.5) $$ \begin{align} \mu(\operatorname{supp}(z)\backslash(A_1^+\cup A_2^+))=0, \end{align} $$

(2.6) $$ \begin{align} |\lambda_k|=\int_{A_k}|z(t)|\,d\mu=\int_{A_k^+}|z(t)|\,d\mu \end{align} $$

and

(2.7) $$ \begin{align} \mu\{t\in A_k: |z(t)|>|x_k(t)|\}=0 \end{align} $$

for $k=1,2$ . Indeed, by (2.1), (2.2) and Lemma 2.2,

(2.8) $$ \begin{align} &\min\{\|f(z)\pm f(x_1)\|\}=\min\{\|(\theta_1\lambda_1\pm1) f(x_1)+\theta_2\lambda_2 f(x_2)\|\} \nonumber\\ &\quad =\min\{1+\lambda_1, 1-\lambda_1\}+|\lambda_2|=1-|\lambda_1|+|\lambda_2|=2(1-|\lambda_1|). \end{align} $$

On the other hand, by (2.4),

(2.9) $$ \begin{align} &\min\{\|f(z)\pm f(x_1)\|\}=\|z-\alpha_1 x_1\| \nonumber\\ & \quad =\int_{A_1\cup A_1^c}|z(t)-\alpha_1 x_1(t)|\,d\mu = \int_{A_1}|z(t)-\alpha_1 x_1(t)|\,d\mu+1-\int_{A_1}|z(t)|\,d\mu \nonumber\\ & \quad = 1+\int_{A_1^+}(||z(t)|-|x_1(t)||-|z(t)|)\,d\mu+\int_{A_1^-}(|z(t)|+|x_1(t)|-|z(t)|)\,d\mu \nonumber\\ & \quad = 1+\int_{A_1^+}(||z(t)|-|x_1(t)||-|z(t)|)\,d\mu+1-\int_{A_1^+}|x_1(t)|\,d\mu \nonumber\\ & \quad = 2-2\int_{A_1^+}\min\{|z(t)|,|x_1(t)|\}\,d\mu\geq2-2\int_{A_1^+}|z(t)|\,d\mu. \end{align} $$

It follows from (2.8) and (2.9) that

$$ \begin{align*}|\lambda_1|\leq \int_{A_1^+}|z(t)|\,d\mu.\end{align*} $$

By the same method, we also obtain

$$ \begin{align*}|\lambda_2|\leq \int_{A_2^+}|z(t)|\,d\mu.\end{align*} $$

Since $\mu (A_1\cap A_2)=0$ , we see that

$$ \begin{align*}1=|\lambda_1|+|\lambda_2|\leq \int_{A_1^+}|z(t)|\,d\mu+ \int_{A_2^+}|z(t)|\,d\mu\leq \|z\|=1.\end{align*} $$

This proves (2.5) and (2.6) and shows that the inequality in (2.9) becomes

$$ \begin{align*} \int_{A_k^+}\min\{|z(t)|,|x_k(t)|\}\,d\mu=\int_{A_k^+}|z(t)|\,d\mu \end{align*} $$

for $k=1,2$ . This, along with (2.5), gives (2.7).

We will now establish (2.3). Without loss of generality, we need only prove that there exists some $\mu _1$ with $|\mu _1|=|\lambda _1|$ such that $z(t)=\mu _1x_1(t)$ for almost every (a.e.) $t\in A_1$ . For every measurable subset $A\subset A_1$ with $\mu (A)>0$ , equations (2.5) and (2.7) imply that

$$ \begin{align*} &\bigg\|\frac{x_1\cdot \chi_A}{\|x_1\cdot \chi_A\|}\pm\alpha_1 z\bigg\|=\int_{A}\bigg|\frac{x_1(t)}{\|x_1\cdot \chi_A\|}\pm\alpha_1 z(t)\bigg|\,d\mu+\int_{A^c}|z(t)|\,d\mu\\ & \quad= \int_{A\cap A_1^+}\bigg(\frac{|x_1(t)|}{\|x_1\cdot \chi_A\|}\pm|z(t)|\bigg)\,d\mu+\int_{A\cap A_1^-}\frac{|x_1(t)|}{\|x_1\cdot \chi_A\|}\,d\mu+1-\int_A|z(t)|\,d\mu, \end{align*} $$

and then

(2.10) $$ \begin{align} &\min\bigg\{\bigg\|\frac{x_1\cdot \chi_A}{\|x_1\cdot \chi_A\|}\pm z\bigg\|\bigg\}=\min\bigg\{\bigg\|\frac{x_1\cdot \chi_A}{\|x_1\cdot \chi_A\|}\pm \alpha_1 z\bigg\|\bigg\} \nonumber\\ & \quad = 1+\int_{A\cap A_1^-}\frac{|x_1(t)|}{\|x_1\cdot \chi_A\|}\,d\mu+\int_{A\cap A_1^+}\bigg(\frac{|x_1(t)|}{\|x_1\cdot \chi_A\|}- |z(t)|-|z(t)|\bigg)\,d\mu \nonumber\\ & \quad = 2-2\int_{A\cap A_1^+}|z(t)|\,d\mu=2-2\int_{A}|z(t)|\,d\mu. \end{align} $$

On the other hand, since f is odd,

(2.11)

It follows from (2.10) and (2.11) that

$$ \begin{align*} \int_{A}(|z(t)|-|\lambda_1x_1(t)|)\,d\mu\geq 0. \end{align*} $$

Since $A\subset A_1$ is arbitrary, $|z(t)|\geq |\lambda _1x_1(t)|$ for a.e. $t\in A_1$ . The fact that

$$ \begin{align*}\int_{A_1}|z(t)|\,d\mu=|\lambda_1|=\int_{A_1}|\lambda_1 x_1(t)|\,d\mu\end{align*} $$

shows that $|z(t)|= |\lambda _1x_1(t)|$ for a.e. $t\in A_1$ , and so $z(t)=\alpha _1|\lambda _1|x_1(t)$ for a.e. $t\in A_1$ by (2.5). We have established (2.3) and this completes the proof.

Proof. Combining Lemma 2.1 with Theorem 2.3,

$$ \begin{align*} \bigg\{f\bigg(\sum_{k=1}^n \frac{\theta_k}{n} x\bigg):|\theta_k|=1, k=1,\ldots, n \bigg\}=\bigg\{\sum_{k=1}^n \frac{\theta_k}{n}f(x_k):|\theta_k|=1, k=1,\ldots, n \bigg\}. \end{align*} $$

It follows that for any choice of signs $\theta _k=\pm 1\ (1\leq k\leq n)$ , there exist some signs $\theta ^{\prime }_k=\pm 1\ (1\leq k\leq n)$ such that

$$ \begin{align*} \bigg\|\sum_{k=1}^n \theta_k f(x_k)\bigg\|=n\bigg\|f\bigg(\sum_{k=1}^n\frac{\theta_k'}{n} x_k\bigg)\bigg\|=n. \end{align*} $$

The conclusion follows from Lemma 2.2.

Proof of Theorem 1.1.

If $\dim L^1(\mu )=1$ , then $L^1(\mu )\cong \mathbb R$ and $\widetilde {f}:\mathbb R\to Y$ is a surjective linear isometry. We may assume that $\dim L^1(\mu )\geq 2$ . The proof will be divided into several steps.

Step I. Set $Z_E:=\{x\in {L^1(\mu )}: \mu (\operatorname {supp}(x)\cap E^c)=0\}$ for $E\in \mathfrak {M}$ . Fix a set $A\in \mathfrak {M}$ such that $Z_A\neq \{0\}$ and $Z_{A^c}\neq \{0\}$ hold simultaneously. Note that this implies that $\mu (A)>0$ and $\mu (A^c)>0$ .

The existence of the above set A can be proved as follows. Fix an $x\in L^1(\mu )\backslash \{0\}$ . Set $A_n=\{t\in \Omega :|x(t)|>1/n\}$ , $n=1,2,\ldots .$ Then $\mu (A_n)<\infty $ for all n. Since $x\neq 0$ , there is an $n_0$ such that $\mu (A_{n_0})>0$ . Let $A=A_{n_0}$ . Then $0<\mu (A)<\infty $ and $Z_A\neq \{0\}$ since $\chi _A\in Z_A$ . If $Z_{A^c}\neq \{0\}$ , we are done. Otherwise, if $Z_{A^c}=\{0\}$ , then $L^1(\mu )=Z_A$ . If there exists a set $A_0\subset A$ such that $\mu (A_0)>0$ and $\mu (A\backslash A_0)>0$ , then $A_0$ is the desired set. Otherwise, if no such $A_0$ exists, then $L^1(\mu )$ consists of only scalar multiples of $\chi _A$ and has dimension $1$ , which contradicts the assumption.

Step II. The unit sphere $S_{L^1(\mu )}$ of $L^1(\mu )$ can be uniquely represented as

$$ \begin{align*} S_{L^1(\mu)}=\{(1-\lambda)z_1+\lambda z_2: z_1\in S_{Z_A}, z_2\in S_{Z_{A^c}}, 0\leq\lambda\leq1\}. \end{align*} $$

Indeed, for each $x\in S_{L^1(\mu )}$ with $\mu (\operatorname {supp}(x)\cap A)>0$ and $\mu (\operatorname {supp}(x)\cap A^c)>0$ ,

$$ \begin{align*} x=\|x\cdot\chi_A\|\cdot\frac{x\cdot\chi_A}{\|x\cdot\chi_A\|}+\|x\cdot\chi_{A^c}\|\cdot\frac{x\cdot\chi_{A^c}}{\|x\cdot\chi_{A^c}\|}. \end{align*} $$

The existence is established. For the uniqueness, assume that

$$ \begin{align*} x=(1-\lambda_1)z_1+\lambda_1z_2=(1-\lambda_2)z^{\prime}_1+\lambda_2z^{\prime}_2, \end{align*} $$

where $z_1,z^{\prime }_1\in S_{Z_A}, z_2,z^{\prime }_2\in S_{Z_{A^c}}, 0\leq \lambda _1,\lambda _2\leq 1$ . It follows that

$$ \begin{align*} (1-\lambda_1)z_1-(1-\lambda_2)z^{\prime}_1=\lambda_2z^{\prime}_2-\lambda_1z_2\in Z_A\cap Z_{A^c}=\{0\}, \end{align*} $$

and then $\lambda _1=\lambda _2$ , $z_1=z^{\prime }_1$ and $z_2=z^{\prime }_2$ .

Step III. For every $z_1\in S_{Z_A}$ , $z_2\in S_{Z_{A^c}}$ and $0\leq \lambda \leq 1$ , Theorem 2.3 implies that there exist $\alpha (z_1,z_2,\lambda ), \beta (z_1,z_2,\lambda )\in \{-1,1\}$ such that

(3.1) $$ \begin{align} f((1-\lambda)z_1+\lambda z_2)=\alpha(z_1,z_2,\lambda)(1-\lambda)f(z_1)+\beta(z_1,z_2,\lambda)\lambda f(z_2). \end{align} $$

The definitions of two extreme cases are added as follows:

(3.2) $$ \begin{align} \alpha(z_1,z_2,1)=\beta(z_1,z_2,0)=\alpha(z_1,z_2,\tfrac12)\beta(z_1,z_2,\tfrac12). \end{align} $$

Write $\gamma (z_1,z_2,\lambda )=\alpha (z_1,z_2,\lambda )\beta (z_1,z_2,\lambda )$ . Define a mapping $g: S_{L^1(\mu )} \to S_Y$ by

(3.3) $$ \begin{align} g((1-\lambda)z_1+\lambda z_2)=\gamma(z_1,z_2,\lambda)(1-\lambda)f(z_1)+\lambda f(z_2). \end{align} $$

Then g is a min-phase-isometry since g is phase-equivalent to f. We claim that

(3.4) $$ \begin{align} g((1-\lambda)z_1+\lambda z_2)=(1-\lambda)g(z_1)+\lambda g(z_2). \end{align} $$

for all $z_1\in S_{Z_A}$ , $z_2\in S_{Z_{A^c}}$ and $0\leq \lambda \leq 1$ .

Indeed, from (3.1) it follows that $ \alpha (z_1,z_2,0)=\beta (z_1,z_2,1)=1$ . Combining this with (3.2) and (3.3),

(3.5) $$ \begin{align} g(z_1)=\gamma(z_1,z_2,\tfrac12)f(z_1)\quad \text{and}\quad g(z_2)=f(z_2). \end{align} $$

Hence, to prove (3.4), we need only show that $\gamma (z_1,z_2,\lambda )=\gamma (z_1,z_2,\frac 12)$ for all $z_1\in S_{Z_A}$ , $z_2\in S_{Z_{A^c}}$ and $0<\lambda <1$ with $\lambda \neq \frac 12$ . If there are such $z_1$ , $z_2$ and $\lambda $ for which $\gamma (z_1,z_2,\lambda )=-\gamma (z_1,z_2,\frac 12)$ , then by Lemma 2.4,

(3.6) $$ \begin{align} &\min\bigg\{\bigg\|g\bigg(\frac{z_1+z_2}2\bigg)\pm g((1-\lambda)z_1+\lambda z_2)\bigg\|\bigg\} \nonumber\\ & \quad = \min\bigg\{\bigg\|\bigg[\gamma\bigg(z_1,z_2,\dfrac12\bigg)\dfrac12\pm\gamma(z_1,z_2,\lambda)(1-\lambda)\bigg]f(z_1)+ \bigg(\dfrac12\pm\lambda\bigg) f(z_2)\bigg\|\bigg\} \nonumber\\ & \quad = \min\bigg\{\bigg|\dfrac12\mp(1-\lambda)\bigg|+\bigg|\dfrac12\pm\lambda\bigg| \bigg\}=\min\bigg\{\dfrac12+\lambda,\dfrac32-\lambda \bigg\}+\bigg|\dfrac12-\lambda\bigg| \nonumber\\ & \quad> |1-2\lambda|=\min\bigg\{\bigg\|\dfrac{z_1+z_2}2\pm ((1-\lambda)z_1+\lambda z_2)\bigg\|\bigg\}, \end{align} $$

which contradicts the fact that g is a min-phase-isometry.

Step IV. The mapping $g: S_{L^1(\mu )} \to S_Y$ is odd.

Indeed, fix $z_1\in S_{Z_A}$ , $z_2\in S_{Z_{A^c}}$ and $0\leq \lambda \leq 1$ . By (3.1) and oddness of f,

$$ \begin{align*} &\alpha(-z_1,-z_2,\tfrac12)f(z_1)+\beta(-z_1,-z_2,\tfrac12)f(z_2)=-2f(\tfrac12(-z_1)+\tfrac12(-z_2))\\ & \quad = 2f(\tfrac12z_1+\tfrac12z_2)=\alpha(z_1,z_2,\tfrac12)f(z_1)+\beta(z_1,z_2,\tfrac12)f(z_2). \end{align*} $$

It follows that $\alpha (-z_1,-z_2,\frac 12)=\alpha (z_1,z_2,\frac 12)$ and $\beta (-z_1,-z_2,\frac 12)=\beta (z_1,z_2,\frac 12)$ . Then (3.3) implies that

$$ \begin{align*} &g(-(1-\lambda)z_1-\lambda z_2)=\gamma(-z_1,-z_2,\tfrac12)(1-\lambda)f(-z_1)+\lambda f(-z_2)\\ & \quad = -\gamma(z_1,z_2,\tfrac12)(1-\lambda)f(z_1)-\lambda f(z_2)=-g((1-\lambda)z_1+\lambda z_2). \end{align*} $$

Step V. The mapping $g: S_{L^1(\mu )} \to S_Y$ is surjective.

Indeed, for any $y\in S_Y$ , since f is surjective and g is phase-equivalent to f, there exists an $x\in S_{L^1(\mu )}$ such that $y=f(x)=\pm g(x)$ . Hence, the oddness of g implies that either $g(x)=y$ or $g(-x)=y$ .

Step VI. The mapping g is an isometry on $S_{Z_A}$ and $S_{Z_{A^c}}$ , respectively.

Indeed, choose arbitrarily $z^{\prime }_1,z^{\prime \prime }_1\in S_{Z_A}$ and fix some $v\in S_{Z_{A^c}}$ . By (3.4),

$$ \begin{align*} &\dfrac13\|g(z^{\prime}_1)-g(z^{\prime\prime}_1)\|=\min\bigg\{\bigg\|g \bigg(\frac{z^{\prime}_1+2v}3\bigg)\pm g\bigg(\frac{z^{\prime\prime}_1+2v}3\bigg)\bigg\|\bigg\}\\ & \quad = \min\bigg\{\bigg\|\frac{z^{\prime}_1+2v}3\pm\frac{z^{\prime\prime}_1+2v}3\bigg\|\bigg\}=\dfrac13\|z^{\prime}_1-z^{\prime\prime}_1\|. \end{align*} $$

This implies that g is an isometry on $S_{Z_A}$ . Similarly, g is an isometry on $S_{Z_{A^c}}$ .

Step VII. For any finite collection $\{x_k\}_{k=1}^n \subset S_{L^1(\mu )}$ with pairwise disjoint supports and for any scalars $\{\lambda _k\}_{k=1}^n$ with $\sum _{k=1}^n|\lambda _k|=1$ ,

(3.7) $$ \begin{align} g\bigg(\sum_{k=1}^n \lambda_kx_k\bigg) = \sum_{k=1}^n \lambda_k g(x_k). \end{align} $$

To verify this, write $x_k=(1-\alpha _k)x_{k1}+\alpha _kx_{k2}$ with $x_{k1}\in S_{Z_A}$ , $x_{k2}\in S_{Z_{A^c}}$ and $0\leq \alpha _k\leq 1$ for $k=1,2,\ldots , n$ . Theorem 2.3 and (3.4) imply that

$$ \begin{align*} g\bigg(\sum_{k=1}^n \lambda_kx_k\bigg) = \sum_{k=1}^n \lambda_k' g(x_k)=\sum_{k=1}^n \lambda_k' ((1-\alpha_k)g(x_{k1})+\alpha_kg(x_{k2})), \end{align*} $$

where $|\lambda _k'|=|\lambda _k|$ for all $k=1,2,\ldots ,n.$ For each fixed k, assume that $1-\alpha _k\neq 0$ (the case $1-\alpha _k=0$ follows by symmetry, as $\alpha _k\neq 0$ in that scenario). By Lemma 2.4 and the fact that g acts as an isometry on $S_{Z_A}$ ,

$$ \begin{align*} &1-(1-\alpha_k)\lambda_k'+1-|(1-\alpha_k)\lambda_k'|=\bigg\|g\bigg(\sum_{j=1}^n \lambda_jx_j\bigg)-g(x_{k1})\bigg\|\\ & \quad = \bigg\|\sum_{j=1}^n \lambda_j x_j-x_{k1}\bigg\|=1-(1-\alpha_k)\lambda_k+1-|(1-\alpha_k)\lambda_k|, \end{align*} $$

which ensures that $\lambda _k'=\lambda _k$ .

Step VIII. The mapping $\widetilde {g}:L^1(\mu )\to Y$ is a surjective linear isometry.

Indeed, since g is surjective and odd, it is easy to see that $\widetilde {g}$ is surjective and homogeneous. To show that $\widetilde {g}$ is a surjective linear isometry, it suffices to prove that $\widetilde {g}$ is additive. Choose $x_1,x_2\in L^1(\mu )\backslash \{0\}$ arbitrarily with $x_1+x_2\neq 0$ . If $x_1$ and $x_2$ have disjoint supports, then (3.7) implies that

$$ \begin{align*} \widetilde{g}(x_1+x_2)=\|x_1+x_2\|g\bigg(\frac{\|x_1\|}{\|x_1+x_2\|}\frac{x_1}{\|x_1\|}+\frac{\|x_2\|}{\|x_1+x_2\|}\frac{x_2}{\|x_2\|}\bigg)=\widetilde{g}(x_1)+\widetilde{g}(x_2). \end{align*} $$

Otherwise, we first deal with simple functions in $Z_A$ . Let

$$ \begin{align*} s_1=\sum_{i=1}^n\alpha_i\chi_{A_i}\quad \text{and}\quad s_2=\sum_{j=1}^m\beta_j\chi_{B_j}, \end{align*} $$

where $\{A_i\}_{i=1}^n$ and $\{B_j\}_{j=1}^m$ are two disjoint collections contained in $\mathfrak {M}$ with $\bigcup _{i=1}^nA_i=\bigcup _{j=1}^mB_j=A$ and $\alpha _i,\beta _j\in \mathbb R$ for $i=1,\ldots ,n$ and $j=1,\ldots ,m$ . Then

$$ \begin{align*} \widetilde{g}(s_1+s_2) & =\widetilde{g}\bigg(\sum_{i=1}^n\sum_{j=1}^m(\alpha_i+\beta_j)\chi_{A_i\cap B_j}\bigg)=\sum_{i=1}^n\sum_{j=1}^m(\alpha_i+\beta_j)\widetilde{g}(\chi_{A_i\cap B_j})\\ & = \sum_{i=1}^n\alpha_i\widetilde{g}(\chi_{A_i})+\sum_{j=1}^m\beta_j\widetilde{g}(\chi_{B_j})=\widetilde{g}(s_1)+\widetilde{g}(s_2). \end{align*} $$

Next, we assume that $x_1,x_2\in Z_A\backslash \{0\}$ with $x_1+x_2\neq 0$ . There are two sets $\{s^{\prime }_n\}_{n=1}^\infty ,\{s^{\prime \prime }_n\}_{n=1}^\infty \subset Z_A$ of simple functions with $\|s^{\prime }_n+s^{\prime \prime }_n\|=\|x_1+x_2\|$ such that $\lim _{n\to \infty }s^{\prime }_n=x_1$ and $\lim _{n\to \infty }s^{\prime \prime }_n=x_2$ . Since g is an isometry on $S_{Z_A}$ ,

$$ \begin{align*} &\widetilde{g}(x_1+x_2)=\|x_1+x_2\|g\bigg(\frac{x_1+x_2}{\|x_1+x_2\|}\bigg)=\lim_{n\to\infty}\|x_1+x_2\|g\bigg(\frac{s^{\prime}_n+s^{\prime\prime}_n}{\|x_1+x_2\|}\bigg)\\ & \quad = \lim_{n\to\infty}\|x_1+x_2\|\bigg(\widetilde{g}\bigg(\frac{s^{\prime}_n}{\|x_1+x_2\|}\bigg)+\widetilde{g}\bigg(\frac{s^{\prime\prime}_n}{\|x_1+x_2\|}\bigg)\bigg) =\lim_{n\to\infty}(\widetilde{g}(s^{\prime}_n)+\widetilde{g}(s^{\prime\prime}_n))\\ & \quad = \lim_{n\to\infty}\bigg(\|s^{\prime}_n\|g\bigg(\frac{s^{\prime}_n}{\|s^{\prime}_n\|}\bigg)+\|s^{\prime\prime}_n\|g\bigg(\frac{s^{\prime\prime}_n}{\|s^{\prime\prime}_n\|}\bigg)\bigg) =\|x_1\|g\bigg(\frac{x_1}{\|x_1\|}\bigg)+\|x_2\|g\bigg(\frac{x_2}{\|x_2\|}\bigg)=\widetilde{g}(x_1)+\widetilde{g}(x_2). \end{align*} $$

Thus, $\widetilde {g}:Z_A\to Y$ is additive. Similarly, $\widetilde {g}:Z_{A^c}\to Y$ is additive. Finally, we assume that $x_1,x_2\in L^1(\mu )\backslash \{0\}$ with $x_1+x_2\neq 0$ . Then

$$ \begin{align*} \widetilde{g}(x_1+x_2) & =\widetilde{g}(x_1\cdot\chi_A+x_2\cdot\chi_A+x_1\cdot\chi_{A^c}+x_2\cdot\chi_{A^c})\\ & = \widetilde{g}(x_1\cdot\chi_A+x_2\cdot\chi_A)+\widetilde{g}(x_1\cdot\chi_{A^c}+x_2\cdot\chi_{A^c})\\ & = \widetilde{g}(x_1\cdot\chi_A)+\widetilde{g}(x_2\cdot\chi_A)+\widetilde{g}(x_1\cdot\chi_{A^c})+\widetilde{g}(x_2\cdot\chi_{A^c})\\ & = \widetilde{g}(x_1\cdot\chi_A+x_1\cdot\chi_{A^c})+\widetilde{g}(x_2\cdot\chi_A+x_2\cdot\chi_{A^c})=\widetilde{g}(x_1)+\widetilde{g}(x_2). \end{align*} $$

Step IX. The mapping g is a surjective isometry. This follows from the previous step and completes the proof.