Verified answer

I'm assuming f is bounded. Let r > 0 be such that |f(t)| < r for all t in [a,b]. ϕ is continuous on [a,b], so it is uniformly continuous on [a,b]. Hence, given e > 0, there corresponds a d, with 0 < d < e/(2r + b - a), such that for all u, v ∈ [m,M], |u - v| < d implies |ϕ(u) - ϕ(v)| < e/(2r + b - a). Since f is Riemann integrable on [a,b], there exists a partition P = {x_0,x_1,...,x_n} of [a,b] such that

U(P,f) - L(P,f) = ∑(i = 1,n) (M_i - m_i) Δx_i < d^2. (*)

Let

M_i(ϕ) = max{ϕ(f(x)): x_(i-1) ≤ x ≤ x_i},

m_i(ϕ) = min{ϕ(f(x)): x_(i-1) ≤ x_i ≤ x_i},

where 1 ≤ i ≤ n. Let S be the set of indices i ∈ {1,2,..,,n} such that M_i - m_i < d; let T be the set of indices i ∈ {1,2,..,n} such that M_i - m_i ≥ d. For all i ∈ S,

M_i(ϕ) - m_i(ϕ) < e/(2r + b - a),

and for all i ∈ T,

M_i(ϕ) - m_i(ϕ) < 2r.

Hence

U(P, ϕ ◦ f) - L(P, ϕ ◦ f)

= ∑(i = 1,n) [M_i(ϕ) - m_i(ϕ)]Δx_i

= ∑(i ∈ S) [M_i(ϕ) - m_i(ϕ)]Δx_i + ∑(i ∈ T) [M_i(ϕ) - m_i(ϕ)]Δx_i

< ∑(i ∈ S) e/(2r + b - a) * Δx_i + ∑(i ∈ T) 2r * Δx_i

= e/(2r + b - a) ∑(i ∈ S) Δx_i + 2r ∑(i ∈ T) 1/(M_i - m_i) * (M_i - m_i) Δx_i

≤ e/(2r + b - a) ∑(i ∈ S) Δx_i + 2r ∑(i ∈ T) 1/d * (M_i - m_i) Δx_i

(since M_i - m_i ≥ d for all i ∈ T)

= e/(2r + b - a) ∑(i ∈ S) Δx_i + 2r(1/d) ∑(i ∈ T) (M_i - m_i) Δx_i

< e/(2r + b - a) ∑(i ∈ S) Δx_i + 2r(1/d)(d^2)

(by (*))

= e/(2r + b - a) ∑(i ∈ S) Δx_i + 2rd

≤ e/(2r + b - a) ∑(i = 1,n) Δx_i + 2rd

= e(b - a)/(2r + b - a) + 2rd

< e(b - a)/(2r + b - a) + 2re/(2r + b - a)

(since 0 < d < e/(2r + b - a))

= e.

Since e is arbitrary, ϕ ◦ f is Riemann integrable on [a,b].