For real numbers α< β and γ > 0, show that if g is integrable over [α+γ, β+γ], then $\int_{\alpha}^{\beta} g(t+\gamma)dt = \int_{\alpha+\gamma}^{\beta+\gamma} g(t)dt$
Prove this change of variables formula by successively considering simple functions, bounded measurable functions, nonnegative integrable functions, and general integrable functions.
My proof:
Let g be integrable over [α+γ, β+γ], then $g^+$and $g^-$ are nonegative integrable functions. There exist an increasing sequence $\{φ_n \}$ of nonegative simple functions such that $g^+=lim φ_n $. Now, since $∫_{α+γ}^{β+γ}χ_{[α+γ,β+γ]} (t) =m([α+γ,β+γ])=m([α,β]+γ)=∫_α^βχ_{[α,β]} (t+γ) $ for the given interval, we have that $∫_{α+γ}^{β+γ} φ_n (t)dt =∫_α^βφ_n (t+γ)dt$ for all n. By the Monotone Convergence Theorem , $∫_{α+γ}^{β+γ}g^+ (t)dt = ∫_α^βg^+(t+γ)dt $. Similarry for $g^-$. Thus $∫_α^βg(t+γ)dt = ∫_{α+γ}^{β+γ}g(t)dt$.
Is this correct or it need something else?
You need to prove the equality for $\textit{arbitrary}$ indicator functions, then build the general case as you describe. So, let $E$ be measurable, and using latin letters for convenience, (in what follows $E\cap [a+c,b+c]$ may be empty, but that's ok):
$\int^{b+c}_{a+c}\chi_E(t)dt=m(E\cap[a+c,b+c])=m((E-c)\cap[a,b])=\int^b_a\chi_{E-c}(t)dt=\int^b_a\chi_{E}(t+c)dt.$
The rest of your proof now goes through unchanged.