Prove $\lim_{\epsilon \downarrow 0}\frac{1}{\epsilon}\left(\int_{\epsilon}^1 f(t)\mathrm{d}t - \int_0^{1-\epsilon} g(t) \mathrm{d}t \right)=0$ Let $f$ and $g$ be two continuous functions defined over $[0,1]$ such that
$$ 
\int_0^1 f(t) \,\mathrm{d}t = \int_0^1 g(t) \,\mathrm{d}t \in \mathbb{R}^+,
$$
can we prove that
$$
\lim_{\varepsilon \downarrow 0} \frac{1}{\varepsilon}\left(\int_{\varepsilon}^1 f(t) \,\mathrm{d}t  - \int_0^{1-\varepsilon} g(t) \,\mathrm{d}t \right) = 0.
$$
I tried the following variable change: $u = t/\varepsilon$. It remains to prove that
$$
\lim_{\varepsilon \downarrow 0} \left(\int_{1}^{\frac{1}{\varepsilon}} f(\varepsilon u) \,\mathrm{d}u  - \int_0^{\frac{1}{\varepsilon}-1} g(\varepsilon u) \,\mathrm{d}u \right) = 0,$$
which I cannot prove.
 A: No. Consider $f\equiv 1$ and $g(t)=2t$, then $\int_0^1 f(t) \,\mathrm{d}t = \int_0^1 g(t) \,\mathrm{d}t=1$. For any $\varepsilon\in(0,1)$, we have
$$\int_{\varepsilon}^1 f(t) \,\mathrm{d}t=1-\varepsilon,\qquad \int_0^{1-\varepsilon} g(t) \,\mathrm{d}t=(1-\varepsilon)^2.$$
Hence
$$\frac{1}{\varepsilon}\left(\int_{\varepsilon}^1 f(t) \,\mathrm{d}t  - \int_0^{1-\varepsilon} g(t) \,\mathrm{d}t \right)=\frac1\varepsilon(1-\varepsilon)[1-(1-\varepsilon)]=1-\varepsilon\to1, \ \ \text{as }\varepsilon\to0^+.$$
Indeed, we have
$$
\lim_{\varepsilon \downarrow 0} \frac{1}{\varepsilon}\left(\int_{\varepsilon}^1 f(t) \,\mathrm{d}t  - \int_0^{1-\varepsilon} g(t) \,\mathrm{d}t \right) = g(1)-f(0).
$$
Since $\int_0^1 f(t) \,\mathrm{d}t = \int_0^1 g(t) \,\mathrm{d}t$, we have
\begin{align*}
\int_{\varepsilon}^1 f(t) \,\mathrm{d}t  - \int_0^{1-\varepsilon} g(t) \,\mathrm{d}t&=\int_0^1 f(t)\,\mathrm{d}t-\int_0^{\varepsilon} f(t) \,\mathrm{d}t-\int_0^1 g(t)\,\mathrm{d}t+\int_{1-\varepsilon}^1 g(t) \,\mathrm{d}t\\
&=\int_{1-\varepsilon}^1 g(t) \,\mathrm{d}t-\int_0^{\varepsilon} f(t) \,\mathrm{d}t.
\end{align*}
By L-Hopital's rule and the fundamental theorem of calculus,
$$\lim_{\varepsilon \downarrow 0} \frac{1}{\varepsilon}\int_{1-\varepsilon}^1 g(t) \,\mathrm{d}t=\lim_{\varepsilon \downarrow 0}g(1-\varepsilon)=g(1),$$
$$\lim_{\varepsilon \downarrow 0} \frac{1}{\varepsilon}\int_0^{\varepsilon} f(t) \,\mathrm{d}t=\lim_{\varepsilon \downarrow 0}f(\varepsilon)=f(0).$$
Therefore,
$$
\lim_{\varepsilon \downarrow 0} \frac{1}{\varepsilon}\left(\int_{\varepsilon}^1 f(t) \,\mathrm{d}t  - \int_0^{1-\varepsilon} g(t) \,\mathrm{d}t \right) = g(1)-f(0).
$$
