Let $[a, b]$ be an interval of $\mathbf{R}$. If $\varphi:[a, b] \rightarrow \mathbf{R}$ is continuous, then the function $F:[a, b] \rightarrow \mathbf{R}$ defined by $F(x)=\int_{[a, x]} \varphi(x) d \lambda(x)$, for $x \in[a, b]$, is differentiable on $[a, b],$ and $F^{\prime}=\varphi$.
For completeness, I'll type the full proof:
Let $x \in[a, b],$ and let $h \neq 0$ such that $x+h \in[a, b] .$ Assuming for example that $h>0,$ we have $$ \frac{F(x+h)-F(x)}{h}=\frac{1}{h} \int_{(x, x+h]} \varphi(y) d \lambda(y)=\int_{(0,1]} \varphi(x+h u) d \lambda(u) $$ where (this is the part that I don't understand) the last equality follows from the change of variable $y=x+h u .$ Since $\varphi$ is continuous, when $h \rightarrow 0,$ the function $u \mapsto \varphi(x+h u)$ converges pointwise to $\varphi(x)$ on $[a, b]$. Moreover all these functions are bounded by $|\varphi|$, which is a bounded, hence integrable function on $[a, b] .$ By dominated convergence, we thus have $$ \int_{(0,1]} \varphi(x+h u) d \lambda(u) \underset{h \rightarrow 0}{\longrightarrow} \varphi(x) $$
The question is, how do I justify the above change of variables and the integrals that follow? This is the form of change of variables that I know. I am unable to use this to conclude the above (I suspect $f(\alpha)= x+h\alpha$).
Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and $(\mathcal{Y},\mathcal{B})$ be a measurable space. $f: (\Omega, \mathcal{F}, \mu) \to (\mathcal{Y},\mathcal{B})$ is a measurable map. Let $f_* \mu (A):= \mu(\{ f^{-1}(A)\})$ for $A \in \mathcal{B}$ be a pushed forward measure. Let $\phi: (\mathcal{Y},\mathcal{B}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be measurable. If $\phi$ is non-negative or integrable, then $\int_\Omega \phi \circ f d \mu= \int_\mathcal{Y} \phi d (f_* \mu)$.
In your case, $\ \Omega=(0,1],$$\ \mathcal{Y}=(x,x+h],$$\ \mathcal{F}=\{A\cap(0,1]\,|\,A\in\mathcal{L}\}\ $ and $\mathcal{B}=\{B\cap(x,x+h]\,|\,B\in\mathcal{L}\}\ $, where $\ \mathcal{L}\ $ is the set of Lebesgue measurable subsets of $\ \mathbf{R}\ $, and $\ \mu=\lambda\ $, Lebesgue measure on $\ (\Omega, \mathcal{F})\ $. With $\ f(u)=x+hu\ $, the measure $\ f_*\mu\ $ is simply \begin{align} f_*\lambda(A)&=\lambda\left(f^{-1}(A)\right)\\ &=\lambda\left(\frac{A-x}{h}\right)\\ &=\frac{\lambda(A)}{h}\ . \end{align} That is, $\ f_*\mu\ $ is just Lebesgue measure on $\ (\mathcal{Y},\mathcal{B})\ $, scaled down by a factor of $\ h\ $. Thus, your change of variable formula gives you \begin{align} \int_{(0,1]}\phi(x+hu)d\lambda(u)&=\int_\mathcal{X}\phi\circ f d\mu\\ &=\int_\mathcal{Y}\phi(y)df_*\mu(y)\\ &=\int_{(x,x+h]}\phi(y)\frac{d\lambda(y)}{h}\ . \end{align}