# How to show $\mathbb{E}(X) = \int_0^{\infty} (1-F_X(x)) \, dx$ for a continuous random variable $X \geq 0$ without assuming $f_X$ exists? [duplicate]

If $X$ is a continuous random variable taking non-negative values only, how do I show that $$\mathbb{E}(X)=\int_0^{\infty}[1-F_X(x)]dx$$

It's easy to give a proof if we assume the density function $f_X$ exists (use Fubini's Theorem to interchange the integrations). However, I got stuck in the general case. What if $X$ does not have a density function?

## marked as duplicate by Nate Eldredge, PhoemueX, Did probability 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 27 '15 at 19:51

• @NateEldredge: The definition I know only requires that the distribution of $X$ assigns zero measure to all singletons (i.e. the distribution function is continuous (not necessarily absolutely continuous)). – PhoemueX Sep 27 '15 at 19:09