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I have to show that the Lebesgue measure of $[a, \infty)$ is infinite, on the measure space $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \lambda)$. However I am a little unsure on my proof of this, so I hope someone could verify it.

Let $(I_n)$ be an increasing sequence of intervals in $\mathcal{B}(\mathbb{R})$, such that $I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots$, where $I_1 = [a, b_1]$, so that $a < b_1 <\infty$, and therefore $I_2 = [a, b_2]$, where $a < b_1 < b_2 < \infty$, then we have the following:

$$\lambda([a, \infty)) = \lambda\left(\bigcup _{n\in \mathbb{N}} I_n\right) = \lim_{n\rightarrow \infty} \lambda(I_n) = \infty$$

Am I wrong in assuming that this would be a correct approach? If not, which approach should I use instead?

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    $\begingroup$ You would have to make sure that $b_n \to \infty$ as $n \to \infty$. You could also just use $I_n = [a, a+n]$. $\endgroup$ – m_squared Sep 14 '17 at 17:53
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Your approach works, but as @m_squared pointed out, you have to make sure that $I_n\uparrow [a,\infty)$.

Here's another approach:

$$\lambda([a,\infty)) = \lambda\left([a,a+1)\cup[a+1, a+2)\ldots\right)=\lambda\left(\bigcup_{k=0}^\infty [a+k, a+k+1)\right)=\\=\sum_{k=0}^\infty \lambda([a+k, a+k+1)) = \sum_{k=0}^\infty 1 =\infty$$

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