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For decreasing positive sequences in ${\ell }^1$, write $x\prec y$ if every initial partial sum of $x$ is at most the corresponding partial sum of $y$ and the finite total sums agree. The finiteness requirement matters: equality of two divergent totals interpreted as infinity is too weak to guarantee a doubly stochastic realization of summable-sequence majorization.

Read equality of the two positive series as equality of a finite total $M$, so the sequences are summable. This is the majorization of summable sequences setting. If equality is instead allowed to mean $\infty =\infty$, the assertion is false: the strictly decreasing sequences $x_i=1+1/(i+1)$ and $y_i=2+1/(i+1)$ satisfy every partial-sum inequality and have divergent equal extended totals. But a normalized nonnegative row of any doubly stochastic matrix averages numbers all at least $2$, and cannot give $x_i<2$. We prove the intended finite-total assertion constructively.

Since $y_j\to 0$ and $0<x_1\le y_1$, there is a finite $k$ such that $y_k\ge x_1>y_{k+1}$. PutThe two-coordinate stochastic averaging matrix sends $(y_k,y_{k+1})$ to $(x_1,u)$. Move the first of these values to the front by a finite permutation, leaving the other values in decreasing order. The tail iswhich is decreasing and strictly positive because $y_{k+1}\le u\le y_k$.

We need the residual majorization in order to repeat the construction. For $n<k$, the first $n$ values of $z$ are at least $x_1$, while every target-tail value is at most $x_1$. ThusFor $n\ge k$, the residual initial sum is the old initial sum with $x_1$ removed:The tail totals also agree, both equalling $M-x_1$. Therefore the same step applies to $(x_2,x_3,\dots )$ and $z$.

Iterating, let $P^{(m)}$ be the product of the first $m$ finite permutations and embedded averaging matrices. Each $P^{(m)}$ is a doubly stochastic matrix, each row is a finitely supported sequence, and the first $m$ entries of $P^{(m)}y$ are $x_1,\dots ,x_m$. Later operations act only on tail rows, so row $i$ becomes permanently fixed once $m\ge i$. Define $p_{ij}$ to be that stabilized row. ThenFor every fixed column $j$ and every $m$, its first $m$ stabilized entries sum to at most one, since they are entries of column $j$ in $P^{(m)}$. Consequently $s_j={\sum }_i{p}_{ij}\le 1$. By Tonelli's theorem,All $y_j$ are strictly positive and every summand in the last expression is nonnegative, so $s_j=1$ for every $j$. Thus both the row sums and the column sums are one, andThe finite-total argument is what prevents loss of column mass in the infinite construction. This completes the doubly stochastic realization of summable-sequence majorization.