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Suppose $f$ is a normalised set function and a submodular set function, and $T_1,\dots ,T_t\subseteq T$ cover every element of $T$ exactly $k$ times, counting repetitions in the indexed collection. Then the displayed inequality holds. Neither the monotone set function property nor the nonnegative function property is required.

Proof. Fix an ordering of $T$ and write $a_j=f(P_j)-f(P_{j-1})$ for its successive marginal increments. The chain lower bound for a submodular set function gives $f(T_i)\ge {\sum }_{x_j\in T_i}{a}_j$. Summing counts each $a_j$ exactly $k$ times, and ${\sum }_j{a}_j=f(T)$. Exact coverage matters because the $a_j$ may be negative. For a monotone set function, the same argument also allows each element to occur at least $k$ times.

Use the chain lower bound for a submodular set function. Order $T=\{x_1,\dots ,x_n\}$, set $P_j=\{x_1,\dots ,x_j\}$ with $P_0=\varnothing$, and put $a_j=f(P_j)-f(P_{j-1})$. For each $U\subseteq T$, reveal its elements in this same order. Since $U\cap P_{j-1}\subseteq P_{j-1}$, diminishing returns and the normalised set function property giveApply this to each $T_i$ and interchange the finite sums. Every element occurs exactly $k$ times, soThis is the exact-cover inequality for submodular set functions. Some $a_j$ can be negative: exact multiplicity, rather than merely at least $k$ occurrences, is what makes the argument valid without monotonicity. In fact the nonnegative function hypothesis is unnecessary here. If $T=\varnothing$, every $T_i$ is empty and both sides vanish.