Problem Suppose that $f:[0,1]\rightarrow [0,\infty)$ be a Lebesgue measurable function such that $$\int_{0}^{1} e^{[f(x)]}dx <\infty. $$ Show that $$\lim_{s\rightarrow \infty} e^s m\left( f^{-1}\left([s,\infty)\right) \right) =0$$
My attempt: According to the given condition, it seems I need to put $f(x)$ into $s$, exponent of $e$. But,, I lost my way.
$$e^s m\left( f^{-1}\left([s,\infty)\right) \right)=e^s \int_{0}^{1} \chi_{[s,\infty)}\circ f(x) dx \\=\int_{0}^{1} e^s \chi_{[s,\infty)}(f(x))dx....... $$
Let $$g_s(x) = e^{s}\cdot \chi_{\{x:f(x)>s\}}.$$ for $s\in \mathbb N$. Note that $g_s(x)\to 0$ almost everywhere, and $|g_s(x)|\le e^{f(x)}\in L_1[0,1]$. By the DCT, $$\lim_{s\to\infty}\int_0^1 g_s(s)\,dx=\lim_{s\rightarrow \infty }e^s\cdot m(\{x:f(x)>s)\})=0$$ Technically, this only proves the limit holds as $s\to \infty$ through the natural numbers. To extend this to $s\to\infty$ through the reals, let $s(n)$ be any sequence converging to $\infty$, and apply the same logic to $g_{s(n)}(x)$. Assuming the axiom of dependent choice, the limit existing through all subsequences implies it exists in general.