I'm trying to solve the following exericize (S.J. Taylor - Introduction to measure and Integration Exercize 5.5.10)
The statement to prove is the following:
If $f:\left[a,b\right]\rightarrow \mathbb{R}$ is a continuous function in $\left(a,b\right)$ then:
\begin{equation*} \lim_{y\to t} \frac{1}{y-t}\left[\int_{a}^{y} f\left(x\right)dx - \int_{a}^{t} f\left(x\right)dx\right] = f\left(t\right) \end{equation*} where the integration is with respect the Lebesgue measure.
Here there is my solution but I have some doubts.
I have to prove that $\forall \epsilon > 0$ $\exists \delta >0$ such that if $|y-t|<\delta$ \begin{equation*} \frac{1}{y-t}\left[\int_{t}^{y} f\left(x\right)dx\right] - f\left(t\right) < \epsilon \end{equation*}
If I fix $\epsilon >0$ Since $f$ is continuous then the integrals are finite and since $f$ is continuous at $t$ $\forall \epsilon >0$ $\exists \delta >0$ such that if $|y-t|<\delta$ then $|f\left(y\right)-f\left(t\right)|<\epsilon$.
\begin{equation*} \begin{split} \left|\frac{1}{y-t}\left[\int_{t}^{y} f\left(x\right)dx\right] - f\left(t\right) \right| & = \frac{1}{y-t}\left[\int_{t}^{y} f\left(x\right)dx\right] - \int_{t}^{y}\frac{f\left(t\right)}{y-t}dx \\ &\le \frac{1}{|y-t|}\int_{t}^{y}|f\left(x\right)-f\left(t\right)|dx \le \frac{1}{|y-t|}\int_{t}^{y}\epsilon \le \epsilon \end{split} \end{equation*}
My doubts are:
It is correct to say that: \begin{equation} \int_{a}^{y} f\left(x\right)dx - \int_{a}^{t} f\left(x\right)dx = \int_{t}^{y} f\left(x\right)dx \end{equation}
Are there some conceptual mistakes along the proof?