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Some problems concerning the additive properties of subsets of R are investigated. From a result of G. G . Lorentz in additive number theory, we show that if P is a nonempty perfect subset of R, then there is a perfect set M with Lebesgue measure zero so that P+M = R. In contrast to this, it is shown that (1) if S is a subset of R is concentrated about a countable set C, then A(S+R) = 0, for every closed set P with A(P) = 0 ; (2) there are subsets G, and G s of R both of which are subspaces of R over the field of rationale such that G,n GZ = {0}, G,+G $ = R and A(G,) = A(G,) = 0. Some other results are obtained under various set theoretical conditions . If 2~40 = N,, then there is an uncountable subset X of R concentrated about the rationale such that if A(G) = 0, then A(G+X) = 0 ; if V = L, then X may be taken to be coanalytic. P. Erdős and E . Straus conjectured and G. G. Lorentz proved that if 1 [1 +2-11, 2] where M,, = U {I(i, m„) : i cB„}, 3) for each n, M„ + 1 c (M„ -1/21") v M u (M„+ 1/2'"), 4) for each n, A(M„) ,1, there is a set B of integers, Be [n, n+2k) so that A + B contains all integers in the interval (n+m+k, n+m+2k] with IBI 0 so that for each i, 1, R defined by T((x, y)) _ (x, x+y). Since, 7r 2(T(M x P)), the projection of T(M x P) into the second coordinate, is all of R and the Lebesgue measure of ar 2(T(M x P)) is no more than the linear measure of T(MxP), T(MxP) has infinite linear measure. Noticing that if Eg~ R 2 , the linear measure of T(E) is no more than three times the linear measure of E, it follows that the linear measure of M x P must be infinite . Q.E.D. We note that our proof of the preceding corollary shows that if A is a subset of R such that for every subset G of R with Lebesgue measure zero, A x G has linear measure zero, then A+E has Lebesgue measure zero, for every set E with Lebesgue measure zero . QUESTION . Is the converse of this result also true? THEOREM 3 . Let P be a nonempty perfect subset of R . There is a subset M of R with Some additive properties of sets of real numbers 189 00 Lebesgue measure 0 so that if P = U Xi , then there is some i so that Xi+M = R. i=1 Proof. Let {p"} , be a countable dense subset of P. For each n and m, let M(n, m) be a perfect subset of R with Lebesgue measure zero so that (PC) [p"1/m, pn + 1/m])+M(n, m) = R .