Which of the following sets in $\mathbb R^2 $ have the positive lebesgue measure? Which of the following sets in $\mathbb R^2 $ have the positive lebesgue measure?
For twe sets $A,B \subset \mathbb R^2$,$A+B$={$a+b:a\in A,b\in B $}
1)S={$(x,y):x^2 +y^2=1$}.
2)S={$(x,y):x^2 +y^2<1$}.
3)S={$(x,y):x=y$}+{$(x,y):x=-y$}.
4)S={$(x,y):x=y$}+{$(x,y):x=y$}.
Since lebesgue measure is a non-negative number.So,the above question can be answered by discarding the sets of  lebesgue measure zero.I'm not not able to recognize sets with lebesgue measure zero.
Also,does there exists any characteristic property $P$ such that a set $S$ has property $P$ iff its lebesgue measure is zero?
 A: The main piece of advice I would offer is that the Lebesgue measure on $\mathbb R^2$ is designed to agree with our intuitive notion of area, on any set for which we have an intuitive notion of area.
For example, the set  $$S = \{ (x, y) \in \mathbb R^2 : x^2 + y^2 < 1 \}$$ is a unit circular disk, which intuitively has area $\pi$. Therefore, we would expect its Lebesgue measure to be $\pi$.
Maybe the simplest way to prove that $\mu(S) = \pi$ would be to think of $\mu(S)$ as the integral,
$$ \mu(S) = \int_{-1}^1 dx\left( \int_{-\sqrt{1-x^2}}^{\sqrt{1 - x^2}}dy \right)= \pi.$$
[I'm afraid I can't think of a totally elementary justification for this integral expression. The simplest explanation I can think of is as follows. The set $S$ is measurable with respect to the product measure on $\mathbb R^2 = \mathbb R \times \mathbb R$ (i.e. the product of the one-dimensional Lebesgue measures on each of the $\mathbb R$ factors). This is because $S$ is an open circular disk, which can be constructed as a union of countable many open rectangles, and open rectangles are certainly measurable w.r.t. the product measure. Furthermore, the Lebesgue measure on $\mathbb R^2$ is the completion of the product measure on $\mathbb R^2 = \mathbb R \times \mathbb R$, so the Lebesgue measure of $S$ agrees with its product measure, which is defined as the double integral I wrote down.]
If we simply wish to prove that $\mu(S)$ is strictly positive, not caring about its precise value, we can do as copper.hat suggested: namely, we can observe that the square $$R = [-\tfrac 1 {\sqrt 2}, \tfrac 1 {\sqrt 2}] \times [-  \tfrac 1 {\sqrt 2}, \tfrac 1 {\sqrt 2}]$$ is contained inside $S$. The Lebesgue measure of the square $R$ is the product of the lengths of its sides: $m(R) = \sqrt 2 \times \sqrt 2 = 2$. Hence $m(S) \geq m(R) = 2 > 0$.

The integration technique can be applied on other examples. For example, take $$L = \{ (x,y) \in \mathbb R^2 : x = y \}.$$
This is a straight line, which is infinitesimally thin, so intuitively it has zero area. We therefore expect its Lebesgue measure to be zero.
Its Lebesgue measure is given by the double integral,
$$ \mu(L) = \int_{- \infty}^{\infty} dx \left( \int_{x}^x dy\right) = 0. $$
Alternatively, we can use the standard fact that the Lebesgue measure is invariant under rotations. Rotating $L$ through $45$ degrees, we learn that  the Lebesgue measure of $L$ is the same as the Lebesgue measure of the horizontal line,$$L' = \{ (x,0) : x \in \mathbb R \}.$$ Furthermore, $L'$ is the union of the countable collection of intervals $$E_n = [n, n+1] \times \{ 0 \}, \ \ \ \ n \in \mathbb Z,$$ which each individually have zero Lebesgue measure, so $L'$, and hence $L$ too, have zero Lebesgue measure.
For a third perspective, think of $L$ as the union of two half-lines,
$$L_+ = \{(x,y) : x = y \geq 0 \},  \ \ \ \ \ L_- = \{(x,y) : x = y \leq 0 \}.$$
Observe that it is possible to cover the half-line $L_+$ using a countable collection of squares with sides of length $\epsilon, \frac \epsilon 2, \frac \epsilon 3, \frac \epsilon 4, \dots $, placed corner to corner, for any choice of $\epsilon > 0$. Therefore, the measure of $L_+$ is smaller than the sum of the measures of these squares, which is equal to $\sum_{n \geq 1} \frac {\epsilon^2}{n^2} = \frac{\epsilon^2 \pi^2}{6}$. Since $\epsilon $ is arbitrary, the measure of $L_+$ must be zero. By a similar argument, the measure of $L_-$ is zero too, so $L = L_+ \cup L_-$ has zero measure.
I hope I have managed to provide enough intuition and machinery for you to tackle similar problems in the future.
A: This answer relies (i) on the fact that the measure of a rectangle
is the usual area, (ii) on the fact that if $A \subset B$ are measurable, then $mA \le m B$ and (iii), if $A = \cup_n A_n$ where
the $A_n$ are disjoint, then $mA = \sum_n m A_n$.
For 1,2) note that
$A=[-{1 \over 2}, -{1 \over 2}]^2 \subset S$ and $mA = 1$ hence $mS \ge 1$ and so the measure is positive.
For 3), note that any point $(x,y)$ can be written as 
$({1 \over 2} (x+y), {1 \over 2} (x+y)) + ({1 \over 2} (x-y), -{1 \over 2} (x-y))$ hence $S= \mathbb{R}^2$. Since $A \subset S$ we see that
the measure is positive.
For 4), note that $S= \{ (x,x) \}_{x \in \mathbb{R}}$.
Let $H_n =1 + {1 \over 2} + \cdots + {1 \over n}$ and $H_0 = 0$
for convenience, and choose $\epsilon>0$.
Let $S_\epsilon = \cup_{n=0}^\infty ( (-\epsilon H_{n+1}, -\epsilon H_n]^2  \cup [\epsilon H_n, \epsilon H_{n+1})^2 )$ and note
that $S \subset S_\epsilon$ for all $\epsilon>0$ and that
$m S_\epsilon = 2\epsilon^2 \sum_{n=1}^\infty {1 \over n^2}$. Since
the summation is finite, and $\epsilon>0$ is arbitrary, we see that
$m S = 0$.
