Complex l one functionals on finite measure spaces have bounded densities

Restrict $\mathrm{Re}\mathrm{\Lambda }$ and $\mathrm{Im}\mathrm{\Lambda }$ to real $L^1$ classes. They are real-linear with norm at most $M$. F1 with $p=1$, under F3, gives real $h_1,h_2\in L^1$ representing them on bounded real measurable functions.

Fix either density $h$ and its real functional $A$. For $k\ge 1$, set $E_k=\{h>M+1\mathrm{/}k\}$. Its indicator is in $L^1$ since the measure space is finite. Then $(M+1\mathrm{/}k)\mu (E_k)\le {\int }_{E_k}h=A(1_{E_k})\le M\mu (E_k)$, so $\mu (E_k)=0$. Apply the same reasoning to $\{-h>M+1\mathrm{/}k\}$ with the negative functional. Their countable union shows $\mathrm{\mid }h\mathrm{\mid }\le M$ almost everywhere. Define $g=h_1+i{h}_2$; it is measurable and $\mathrm{\mid }g\mathrm{\mid }\le 2M$ almost everywhere. We may set it to zero on the explicitly determined null set where this bound fails.

For a bounded complex function $f=a+ib$ with real bounded $a,b$, complex linearity gives $\mathrm{\Lambda }(f)=\mathrm{\Lambda }(a)+i\mathrm{\Lambda }(b)=\int a(h_1+i{h}_2)+i\int b(h_1+i{h}_2)=\int fg$, using F2. For arbitrary $f\in L^1$, let $f_N=f\min (1,N\mathrm{/}\mathrm{\mid }f\mathrm{\mid })$, with zero value at $f=0$. Then $f_N$ is bounded, $f_N\to f$ pointwise and $\mathrm{\mid }{f}_N-f\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$, so F4 gives $\mathrm{\parallel }{f}_N-f{\mathrm{\parallel }}_1\to 0$. Boundedness of $\mathrm{\Lambda }$ gives $\mathrm{\Lambda }(f_N)\to \mathrm{\Lambda }(f)$, while step 2.1 and F5 give $\mathrm{\mid }\int (f_N-f)g\mathrm{\mid }\le 2M\mathrm{\parallel }{f}_N-f{\mathrm{\parallel }}_1\to 0$. Passing to the limit proves the formula for all $L^1$ classes.

If $M=0$, the same level-set argument gives $g=0$ almost everywhere; if $\mu (X)=0$, $L^1$ is the zero space and take $g=0$. Indicator tests were used only on a finite-measure space, and truncations were explicitly defined. The full AC use is precisely F1, not a presumed complex duality theorem. $\mathrm{\square }$