# Limit of Integral of square of function

Let $f: [0,1)\rightarrow \mathbb{R}$ be continuous and nonnegative on $[0,1)$. Prove that if $\lim_{a\rightarrow 1^{-}}{\int_{0}^{a}{f(x)^{2}dx}}$ exists, then $\lim_{a\rightarrow 1^{-}}{\int_{0}^{a}{f(x)dx}}$ exists too.

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What have you tried? –  Andrew Aug 19 '12 at 18:07

The Cauchy-Schwarz inequality (taking the inner product with the function $1$) shows that if $g \in L^2[0,1)$, then $|\int_0^1 g(x) dx | \leq \sqrt{\int_0^1 |g(x)|^2 dx}$.

Then consider the functon $f_a(x) = f(x) 1_{[0,a]} (x)$, with $a \in [0,1)$. Since $f$ is non-negative, $\int_0^a f(x) dx \leq \sqrt{\int_0^a |f(x)|^2 dx} \leq \lim_{a\rightarrow 1^{-}} \sqrt{\int_0^a |f(x)|^2 dx}$.

Since $a \mapsto \int_0^a f(x) dx$ is non-decreasing, it follows that the limit exists.

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Since $f$ is nonnegative $g(a)= \int_{0}^{a} f(x) \mathrm{d}x$ in increasing so it suffices to bound $g(x) \leq M \,\,\forall x \in [0,1)$. Now since $f$ is continuous $M_a=\{x \in [0,a)|f(x) \leq 1 \}$ is a measurable set so you can integrate $g$ there. So is $L_a=[0,a) / M_a$.

Now we have $g(a)= \int _{x \in M_a} f(x) \mathrm{d}x+ \int _{x \in L_a} f(x) \mathrm{d}x \leq 1 \cdot |M_a| + \int_{0}^{a} f(x)^2 \mathrm{d}x \leq 1 + \lim_{a \rightarrow 1}\int_{0}^{a} f(x)^2 \mathrm{d}x$.

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The following argument lacks rigor and could probably be flushed out quite a bit, but I think it's the simplest approach. Define $g:[0,1) \to \mathbb{R}$ such that $g(x) = 1$ when $f(x) < 1$ and $g(x) = f(x)$ when $f(x) \ge 1$. Then the existance of $\lim_{a\rightarrow 1^{-}}{\int_{0}^{a}{f(x)^{2}dx}}$ implies the existance of $\lim_{a\rightarrow 1^{-}}{\int_{0}^{a}{g(x)^{2}dx}}$. You could then reason that because $\lim_{a\rightarrow 1^{-}}{\int_{0}^{a}{f(x)dx}}$ is bounded above by $\lim_{a\rightarrow 1^{-}}{\int_{0}^{a}{g(x)^{2}dx}}$, and because $\int_{0}^{a}{f(x)dx}$ is a monotonically increasing function, the integral in question must have a supremum on $[0,1)$ and therefore the limit must exist.

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