# The existence of a measurable set with measure between rationals and the reals [duplicate]

Is there a measurable subset E ⊆ R such that whenever a < b are real numbers we have both $m(E ∩ (a, b)) > 0$ and $m((a, b) -E) > 0$ ? This is an extra question on my real analysis class, but it is not graded, just for fun.

I was thinking about the "fat Cantor set"(https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set), and play with it a little bit, but somehow I can still find "gaps" between real numbers. Can anyone give me some hint so I can carry on ? Thanks !!!

## marked as duplicate by Jack D'Aurizio real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 26 '15 at 20:22

• Thanks ! !!!!!!! – starry1990 Sep 27 '15 at 19:00

As a slight rephrasing of the problem, let's try to construct a pair of disjoint measurable sets $A, B\subset\mathbb{R}$ such that both $A\cap(a,b)$ and $B\cap(a,b)$ have positive measure for any interval $(a,b)$. Here's the idea behind one possible solution: it suffices to consider intervals $(a,b)$ where $a$ and $b$ are both rational. Let $A$ and $B$ take turns picking fat Cantor sets they want to contain (that are disjoint from all the fat Cantor sets chosen so far). Since there are only countably many rational intervals, both $A$ and $B$ can make sure that they eventually choose a fat Cantor set contained in each rational interval.