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Construct a $\sigma$-algebra $A$ of subsets of $\mathbb{R}$ such that no open interval is measurable with respect to $A$, although any singleton $\{x\}$ is ($x \in \mathbb{R}$).

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You will want all $\{ x \}$ to be in $A$, for $x \in \mathbb{R}$.

Then all countable sets have to be in $A$. And their complements.

This suggests taking $$ A = \{ S \subseteq \mathbb{R} : \text{$S$ or $\mathbb{R} \setminus S$ is countable} \}. $$

Now please verify that this is a $\sigma$-algebra, and that it contains no open interval besides $\mathbb{R}$.

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    do you mean that any interval in R has uncountable complement, but how we can approve it. – 2017-02-17
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    Do you know that $\mathbb{R}$ is uncontable? – 2017-02-17
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    yes i know that but it ok to say that since R is uncountable therefore no open interval is countable or has countable complement. – 2017-02-17
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    Do you know that there is a bijection between any open interval and $\mathbb{R}$? – 2017-02-17
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    let B in A that means B in R, suppose that B is countable so, R\B is uncountable, and since R is uncountable which is covered R\B , then B complement ⊆ R , and B complement in A. Is this correct way to approve sigma algebra? – 2017-02-19