On the product of two sigma-finite measure spaces, Tonelli's theorem states that a nonnegative measurable function has measurable iterated-integral functions andwith values allowed to be infinite. Fubini's theorem states that if $f$ is measurable and ${\int }_{X\times Y}|f|<\infty$, then its sections are integrable almost everywhere, both iterated integrals exist as finite integrals, and the same equality holds, also for real or complex signed functions. Tonelli applied to $|f|$ can first be used to verify the absolute-integrability hypothesis of Fubini.

Let $S_N={\sum }_{k=0}^N{a}_k$. These are nonnegative measurable functions increasing to the extended-valued sum $S$. Monotone convergence givesA nonnegative function with finite integral cannot be infinite on a set of positive measure. Hence $S(x)<\infty$ almost everywhere, which is exactly almost-everywhere convergence of the nonnegative series. One may redefine the sum arbitrarily on the exceptional null set to obtain an ordinary integrable representative.

An invertible matrix is exactly an ordered basis of its column space. The first column has $q^n-1$ choices. After $j$ independent columns have been chosen, their span has $q^j$ elements, so the next column has $q^n-q^j$ choices. Therefore

At the current dual prices $(1,1)$, the new product has reduced cost $5-(1+2)=2>0$, so it can increase profit. Its transformed resource column isIncreasing $x_4$ by $t$ changes the old basic variables to $x_1=9+7t$ and $x_3=3-3t$. Thus $x_3$ leaves at $t=1$, givingFor a certificate, take dual prices $(1/3,7/3)$. They price the four products at $3,17/3,26/3,5$, respectively, each at least its profit. Their resource total is $33/3+18(7/3)=53$, equal to the achieved profit. Weak duality proves optimality, and the profit has increased by two.

Introduce nonnegative slack variables $s_1,s_2$ and use a simplex tableau with the last row representing $z-c^{\mathsf{T}}x=0$. The starting tableau isEnter $x_3$ and let $s_2$ leave. Dividing the second row by three and eliminating its column givesEnter $x_1$ and let $s_1$ leave. The next simplex tableau isThusThe pivot choices and the optimality certificate are justified in the following parts.

Let $x_j\ge 0$ be the production amounts of the three distinct products; the repeated first-product labels in the printed table are interpreted using the surrounding text and the three separate columns. The linear program is

Before a potential update, let $S$ be the reached rows and $T$ the reached columns in an alternating forest grown from unmatched rows. Every reached column is matched, or an augmentation would already have occurred, and its matched row is reached. Each tree has one more reached row than reached column, so $|S|-|T|>0$.

The update amount is the smallest slack across $S\times T^c$. It is strictly positive because all zero-slack edges have already been explored, and the complete assignment problem guarantees that the set is nonempty before a perfect matching is found. Increasing $u_i$ for $i\in S$ and decreasing $v_j$ for $j\in T$ therefore increases the dual sum byAll feasibility inequalities remain valid by the definition of the minimum slack.

Begin with $u=v=0$ and the printed matching $\{(1,1),(3,2),(5,3)\}$. Grow an alternating tree from unmatched row two. Its tight edge reaches column two, whose matched row is three. Thus the reached row set is $S=\{2,3\}$ and the reached column set is $T=\{2\}$. Among unreached columns the reduced costs are $(6,5,4,4)$ from row two and $(3,4,2,3)$ from row three, so their minimum is two.

Increase $u_2,u_3$ by two and decrease $v_2$ by two. The new tight edge $(3,4)$ reaches an unmatched column. Augment along $2\to 2\to 3\to 4$, replacing $(3,2)$ by $(2,2),(3,4)$. The matching becomes $\{(1,1),(2,2),(3,4),(5,3)\}$.

Now grow from unmatched row four. Edge $(4,1)$ reaches matched row one, and edge $(1,5)$ reaches unmatched column five. Augment along $4\to 1\to 1\to 5$, replacing $(1,1)$ by $(4,1),(1,5)$. This yieldsNo further potential update is needed. The final potentials are $u=(0,2,2,0,0)$ and $v=(0,-2,0,0,0)$. Their reduced-cost matrix isEvery entry is nonnegative and every chosen entry is zero. The dual sum is $\sum u_i+\sum v_j=2$, so part (a) certifies optimality.