Editing Openai/6935580c-34e4-8001-9136-7bbdf1499790 (section)

==== You described a linear functional LLL that extends the notion of limit to more sequences. But to move it inside an integral you need one of these to hold: ====
# LLL is continuous on the function space that carries the integrals. Concretely, if the integrals live in XXX and LLL is a bounded linear operator L:X→XL: X \to XL:X→X (or L:L1([0,T];X)→XL: L^1([0,T];X)\to XL:L1([0,T];X)→X continuous), then L(∫0tfn)=∫0tL(fn)L\bigg(\int_0^t f_n\bigg)=\int_0^t L(f_n)L(∫0tfn)=∫0tL(fn) will hold in the appropriate sense (because the integral is a continuous linear map).
# Or you have a dominated-convergence-style property: for each fixed sss, LLL commutes with pointwise limits and you also have a uniform integrable bound allowing you to use dominated convergence on the scalar norms ∥fn−f∥\|f_n-f\|∥fn−f∥. But this is just another way of requiring continuity measured in an L1L^1L1-norm.

If LLL is only a linear algebraic extension of the limit that lacks continuity properties with respect to the Bochner/L¹ topology, then you cannot assert

L(∫0tfn(s) ds)  =  ∫0tL(fn(s)) ds.L\Big(\int_0^t f_n(s)\,ds\Big) \;=\; \int_0^t L\big(f_n(s)\big)\,ds.L(∫0tfn(s)ds)=∫0tL(fn(s))ds.
A counterexample is easy to produce with noncontinuous linear functionals on infinite-dimensional spaces: algebraic (Hamel-basis) linear functionals exist but do not commute with limits or integrals.

So the interchange you questioned is not legitimate unless you add a continuity hypothesis on LLL or you obtain norm convergence of the integrands.