= Solution
{c}
Put $E_n=\sup_{t\in[a,b]}|f(t)-\phi_n(t)|\to0$, so the hypothesis is <uniform convergence>. Every <step function> is bounded and <Riemann integrable>: away from its finitely many partition points it is constant, and those exceptional points can be covered by <intervals> of arbitrarily small total length. A <partition of an interval> containing their boundaries makes the difference of its <upper Darboux sum> and <lower Darboux sum> arbitrarily small.
Choose one $n$ with $E_n<\infty$. Boundedness of $\phi_n$ and $|f|\leq|\phi_n|+E_n$ show that $f$ is a <bounded function>. Assume $a<b$, since for $a=b$ all the <integrals> are zero. Given $\eta>0$, choose $n$ so that $2E_n(b-a)<\eta/2$, and then a <partition of an interval> $P$ such that $U(\phi_n,P)-L(\phi_n,P)<\eta/2$. On every partition subinterval, $|f-\phi_n|\leq E_n$ gives
$$
U(f,P)\leq U(\phi_n,P)+E_n(b-a),\qquad L(f,P)\geq L(\phi_n,P)-E_n(b-a).
$$
Thus
$$
U(f,P)-L(f,P)\leq U(\phi_n,P)-L(\phi_n,P)+2E_n(b-a)<\eta.
$$
The <Riemann integrability criterion> proves that $f$ is <Riemann integrable>. Finally, <linearity of the integral> and <monotonicity of the Riemann integral> give
$$
\left|\int_a^bf(t)\,dt-\int_a^b\phi_n(t)\,dt\right|\leq\int_a^b|f(t)-\phi_n(t)|\,dt\leq(b-a)E_n\longrightarrow0.
$$
Therefore
$$
\boxed{\int_a^bf(t)\,dt=\lim_{n\to\infty}\int_a^b\phi_n(t)\,dt.}
$$
This proves the relevant case of <uniform convergence preserves integrals on a compact interval>.
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