proof of Barbalat’s lemma

We suppose that $y(t)\ot\\rightarrow 0$ as $t\\rightarrow\\infty$ . There exists a sequence $(t_{n})$ in $\\mathbb{R}_{+}$ such that $t_{n}\\rightarrow\\infty$ as $n\\rightarrow\\infty$ and $|y(t_{n})|\\geq\\varepsilon$ for all $n\\in\\mathbb{N}$ . By the uniform continuity of $y$ , there exists a $\\delta>0$ such that, for all $n\\in\\mathbb{N}$ and all $t\\in\\mathbb{R}$ ,

|t_{n}-t|\\leq\\delta\\;\\Rightarrow\\;|y(t_{n})-y(t)|\\leq\\frac{\\varepsilon}{2}.

So, for all $t\\in[t_{n},t_{n}+\\delta]$ , and for all $n\\in\\mathbb{N}$ we have

	$\\displaystyle\|y(t)\|$	$\\displaystyle=$	$\\displaystyle\|y(t_{n})-(y(t_{n})-y(t))\|\\geq\|y(t_{n})\|-\|y(t_{n})-y(t)\|\\geq$
		$\\displaystyle\\geq$	$\\displaystyle\\varepsilon-\\frac{\\varepsilon}{2}=\\frac{\\varepsilon}{2}.$

Therefore,

\\left|\\int_{0}^{t_{n}+\\delta}y(t)dt-\\int_{0}^{t_{n}}y(t)dt\\right|=\\left|\\int_{%t_{n}}^{t_{n}+\\delta}y(t)dt\\right|=\\int_{t_{n}}^{t_{n}+\\delta}|y(t)|dt\\geq%\\frac{\\varepsilon\\delta}{2}>0

for each $n\\in\\mathbb{N}$ . By the hypothesis, the improprer Riemann integral $\\int_{0}^{\\infty}y(t)dt$ exists, and thus the left hand side of the inequality converges to 0 as $n\\rightarrow\\infty$ , yielding a contradiction.